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July 13, 2025By
खण्ड ‘ A ′ A ′ A^(‘)A^{\prime}A′ SECTION ‘ A ′ A ′ A^(‘)A^{\prime}A′
1.(a) मान लीजिए कि S 3 S 3 S_(3)S_3S3 व Z 3 Z 3 Z_(3)Z_3Z3 क्रमशः 3 प्रतीकों का क्रमचय समूह एवं मॉड्यूल 3 अवशिष्ट वर्गों के समूह हैं। दर्शाइए कि S 3 S 3 S_(3)S_3S3 का Z 3 Z 3 Z_(3)Z_3Z3 में तुच्छ समाकारिता के अतिरिक्त कोई भी समाकारिता नहीं है ।
Let S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3 be permutation group on 3 symbols and group of residue classes module 3 respectively. Show that there is no homomorphism of S 3 S 3 S_(3)S_3S3 in Z 3 Z 3 Z_(3)Z_3Z3 except the trivial homomorphism.
Answer:
To show that there is no homomorphism of S 3 S 3 S_(3)S_3S3 into Z 3 Z 3 Z_(3)Z_3Z3 except the trivial homomorphism, we can use the following properties of homomorphisms:
  1. A homomorphism ϕ : G → H ϕ : G → H phi:G rarr H\phi: G \to Hϕ:G→H preserves the identity element, i.e., ϕ ( e G ) = e H ϕ ( e G ) = e H phi(e_(G))=e_(H)\phi(e_G) = e_Hϕ(eG)=eH.
  2. A homomorphism ϕ : G → H ϕ : G → H phi:G rarr H\phi: G \to Hϕ:G→H preserves the group operation, i.e., ϕ ( a ∗ b ) = ϕ ( a ) ∗ ϕ ( b ) ϕ ( a ∗ b ) = ϕ ( a ) ∗ ϕ ( b ) phi(a**b)=phi(a)**phi(b)\phi(a \ast b) = \phi(a) \ast \phi(b)ϕ(a∗b)=ϕ(a)∗ϕ(b).
  3. A homomorphism ϕ : G → H ϕ : G → H phi:G rarr H\phi: G \to Hϕ:G→H preserves the order of elements, i.e., if a a aaa has order n n nnn in G G GGG, then ϕ ( a ) ϕ ( a ) phi(a)\phi(a)ϕ(a) has order dividing n n nnn in H H HHH.

Properties of S 3 S 3 S_(3)S_3S3 and Z 3 Z 3 Z_(3)Z_3Z3

  1. S 3 S 3 S_(3)S_3S3 is the permutation group on 3 symbols, and it has 3 ! = 6 3 ! = 6 3!=63! = 63!=6 elements.
  2. Z 3 Z 3 Z_(3)Z_3Z3 is the group of residue classes modulo 3, and it has 3 elements: [ 0 ] , [ 1 ] , [ 2 ] [ 0 ] , [ 1 ] , [ 2 ] [0],[1],[2][0], [1], [2][0],[1],[2].

Steps to Show No Non-Trivial Homomorphism Exists

  1. Identity Element: Any homomorphism ϕ : S 3 → Z 3 ϕ : S 3 → Z 3 phi:S_(3)rarrZ_(3)\phi: S_3 \to Z_3ϕ:S3→Z3 must map the identity element of S 3 S 3 S_(3)S_3S3 (the identity permutation e e eee) to the identity element of Z 3 Z 3 Z_(3)Z_3Z3 ( [ 0 ] [ 0 ] [0][0][0]).
    ϕ ( e ) = [ 0 ] ϕ ( e ) = [ 0 ] phi(e)=[0]\phi(e) = [0]ϕ(e)=[0]
  2. Order of Elements: The order of any element in Z 3 Z 3 Z_(3)Z_3Z3 divides 3. In S 3 S 3 S_(3)S_3S3, we have elements of order 2 (e.g., transpositions) and elements of order 3 (e.g., 3-cycles). If there exists a non-trivial homomorphism ϕ ϕ phi\phiϕ, then it must map elements of S 3 S 3 S_(3)S_3S3 to elements of Z 3 Z 3 Z_(3)Z_3Z3 in such a way that the order of the image divides the order of the original element.
    However, Z 3 Z 3 Z_(3)Z_3Z3 only has elements of order 1 ( [ 0 ] [ 0 ] [0][0][0]) and order 3 ( [ 1 ] , [ 2 ] [ 1 ] , [ 2 ] [1],[2][1], [2][1],[2]). There are no elements of order 2 in Z 3 Z 3 Z_(3)Z_3Z3.
  3. Contradiction: S 3 S 3 S_(3)S_3S3 contains elements of order 2 (transpositions). Any homomorphism ϕ ϕ phi\phiϕ would have to map these elements to an element in Z 3 Z 3 Z_(3)Z_3Z3 whose order divides 2. Since Z 3 Z 3 Z_(3)Z_3Z3 contains no such elements (other than the identity), we reach a contradiction.
Therefore, the only homomorphism that can exist from S 3 S 3 S_(3)S_3S3 to Z 3 Z 3 Z_(3)Z_3Z3 is the trivial homomorphism that maps all elements of S 3 S 3 S_(3)S_3S3 to the identity element [ 0 ] [ 0 ] [0][0][0] in Z 3 Z 3 Z_(3)Z_3Z3.
1.(b) मान लीजिए R R RRR मुख्य गुणजावली प्रान्त है । दर्शाइए कि R R RRR के विभाग-वलय की प्रत्येक गुणजावली, मुख्य गुणजावली है तथा R / P , R R / P , R R//P,RR / P, RR/P,R के अभाज्यगुणजावली P P PPP के लिए मुख्य गुणजावली प्रान्त है ।
Let R R RRR be a principal ideal domain. Show that every ideal of a quotient ring of R R RRR is principal ideal and R / P R / P R//PR / PR/P is a principal ideal domain for a prime ideal P P PPP of R R RRR.
Answer:

Introduction

The problem asks us to prove two things:
  1. Every ideal of a quotient ring R / P R / P R//PR/PR/P is a principal ideal.
  2. If P P PPP is a prime ideal of R R RRR, then R / P R / P R//PR/PR/P is a principal ideal domain (PID).
To prove these statements, we’ll use the properties of principal ideal domains and quotient rings.

Work/Calculations

Part 1: Every ideal of R / P R / P R//PR/PR/P is a principal ideal

Let I / P I / P I//PI/PI/P be an ideal of R / P R / P R//PR/PR/P, where I I III is an ideal of R R RRR containing P P PPP.
Step 1: Show that I I III is a principal ideal in R R RRR
Since R R RRR is a PID, I I III is generated by a single element a a aaa in R R RRR. That is,
I = ( a ) I = ( a ) I=(a)I = (a)I=(a)
Step 2: Show that I / P I / P I//PI/PI/P is generated by a + P a + P a+Pa+Pa+P in R / P R / P R//PR/PR/P
Let’s substitute the values:
I / P = ( a ) + P I / P = ( a ) + P I//P=(a)+PI/P = (a) + PI/P=(a)+P
After substituting, we can see that I / P I / P I//PI/PI/P is generated by a + P a + P a+Pa+Pa+P in R / P R / P R//PR/PR/P.
Therefore, I / P I / P I//PI/PI/P is a principal ideal in R / P R / P R//PR/PR/P.

Part 2: R / P R / P R//PR/PR/P is a PID for a prime ideal P P PPP of R R RRR

Step 1: Show that R / P R / P R//PR/PR/P is an integral domain
Since P P PPP is a prime ideal, R / P R / P R//PR/PR/P is an integral domain.
Step 2: Show that every ideal in R / P R / P R//PR/PR/P is principal
From Part 1, we know that every ideal in R / P R / P R//PR/PR/P is principal.
Step 3: Conclude that R / P R / P R//PR/PR/P is a PID
Since R / P R / P R//PR/PR/P is an integral domain and every ideal in R / P R / P R//PR/PR/P is principal, R / P R / P R//PR/PR/P is a PID.

Conclusion

We have shown that every ideal of a quotient ring R / P R / P R//PR/PR/P is a principal ideal. Additionally, if P P PPP is a prime ideal of R R RRR, then R / P R / P R//PR/PR/P is a principal ideal domain. Both of these statements hold true when R R RRR is a principal ideal domain.
1.(c) सिद्ध कीजिए कि शर्त
| a n + 1 − a n | ⩽ α | a n − a n − 1 | a n + 1 − a n ⩽ α a n − a n − 1 |a_(n+1)-a_(n)| <= alpha|a_(n)-a_(n-1)|\left|a_{n+1}-a_n\right| \leqslant \alpha\left|a_n-a_{n-1}\right||an+1−an|⩽α|an−an−1|, जहाँ पर 0 < α < 1 0 < α < 1 0 < alpha < 10<\alpha<10<α<1 को सभी प्राकृतिक संख्याओं n ⩾ 2 n ⩾ 2 n >= 2n \geqslant 2n⩾2 के लिए सन्तुष्ट करने वाला अनुक्रम ( a n ′ ) a n ′ (a_(n)^(‘))\left(a_n^{\prime}\right)(an′), कॉशी-अनुक्रम होता है ।
Prove that the sequence ( a n ) a n (a_(n))\left(a_n\right)(an) satisfying the condition | a n + 1 − a n | ⩽ α | a n − a n − 1 | , 0 < α < 1 a n + 1 − a n ⩽ α a n − a n − 1 , 0 < α < 1 |a_(n+1)-a_(n)| <= alpha|a_(n)-a_(n-1)|,0 < alpha < 1\left|a_{n+1}-a_n\right| \leqslant \alpha\left|a_n-a_{n-1}\right|, 0<\alpha<1|an+1−an|⩽α|an−an−1|,0<α<1 for all natural numbers n ⩾ 2 n ⩾ 2 n >= 2n \geqslant 2n⩾2, is a Cauchy sequence.
Answer:

Introduction

The problem asks us to prove that a sequence ( a n ) a n (a_(n))\left(a_n\right)(an) satisfying the condition
| a n + 1 − a n | ⩽ α | a n − a n − 1 | , 0 < α < 1 a n + 1 − a n ⩽ α a n − a n − 1 , 0 < α < 1 |a_(n+1)-a_(n)| <= alpha|a_(n)-a_(n-1)|,quad0 < alpha < 1\left|a_{n+1}-a_n\right| \leqslant \alpha\left|a_n-a_{n-1}\right|, \quad 0 < \alpha < 1|an+1−an|⩽α|an−an−1|,0<α<1
for all natural numbers n ⩾ 2 n ⩾ 2 n >= 2n \geqslant 2n⩾2, is a Cauchy sequence. A sequence is said to be Cauchy if for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists an N N NNN such that for all m , n > N m , n > N m,n > Nm, n > Nm,n>N, | a m − a n | < ϵ | a m − a n | < ϵ |a_(m)-a_(n)| < epsilon|a_m – a_n| < \epsilon|am−an|<ϵ.

Work/Calculations

Step 1: Prove that | a n + 1 − a n | a n + 1 − a n |a_(n+1)-a_(n)|\left|a_{n+1} – a_n\right||an+1−an| becomes arbitrarily small

We are given that
| a n + 1 − a n | ⩽ α | a n − a n − 1 | a n + 1 − a n ⩽ α a n − a n − 1 |a_(n+1)-a_(n)| <= alpha|a_(n)-a_(n-1)|\left|a_{n+1}-a_n\right| \leqslant \alpha\left|a_n-a_{n-1}\right||an+1−an|⩽α|an−an−1|
Let’s substitute the values:
| a n + 1 − a n | ⩽ α n | a 2 − a 1 | a n + 1 − a n ⩽ α n a 2 − a 1 |a_(n+1)-a_(n)| <= alpha ^(n)|a_(2)-a_(1)|\left|a_{n+1}-a_n\right| \leqslant \alpha^n \left|a_2 – a_1\right||an+1−an|⩽αn|a2−a1|
After substituting, we see that as n n nnn becomes large, α n α n alpha ^(n)\alpha^nαn approaches zero (since 0 < α < 1 0 < α < 1 0 < alpha < 10 < \alpha < 10<α<1), making | a n + 1 − a n | a n + 1 − a n |a_(n+1)-a_(n)|\left|a_{n+1}-a_n\right||an+1−an| arbitrarily small.

Step 2: Prove that ( a n ) a n (a_(n))\left(a_n\right)(an) is a Cauchy sequence

To prove that ( a n ) a n (a_(n))\left(a_n\right)(an) is a Cauchy sequence, we need to show that for any ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists an N N NNN such that for all m , n > N m , n > N m,n > Nm, n > Nm,n>N, | a m − a n | < ϵ | a m − a n | < ϵ |a_(m)-a_(n)| < epsilon|a_m – a_n| < \epsilon|am−an|<ϵ.
Consider m > n m > n m > nm > nm>n and m , n > N m , n > N m,n > Nm, n > Nm,n>N. We have:
| a m − a n | = | a m − a m − 1 + a m − 1 − a m − 2 + … + a n + 1 − a n | ≤ | a m − a m − 1 | + | a m − 1 − a m − 2 | + … + | a n + 1 − a n | ≤ α m − 1 | a 2 − a 1 | + α m − 2 | a 2 − a 1 | + … + α n | a 2 − a 1 | = α n | a 2 − a 1 | ( 1 + α + α 2 + … + α m − n − 1 ) = α n | a 2 − a 1 | 1 − α m − n 1 − α < α n | a 2 − a 1 | 1 1 − α | a m − a n | = | a m − a m − 1 + a m − 1 − a m − 2 + … + a n + 1 − a n | ≤ | a m − a m − 1 | + | a m − 1 − a m − 2 | + … + | a n + 1 − a n | ≤ α m − 1 | a 2 − a 1 | + α m − 2 | a 2 − a 1 | + … + α n | a 2 − a 1 | = α n | a 2 − a 1 | ( 1 + α + α 2 + … + α m − n − 1 ) = α n | a 2 − a 1 | 1 − α m − n 1 − α < α n | a 2 − a 1 | 1 1 − α {:[|a_(m)-a_(n)|=|a_(m)-a_(m-1)+a_(m-1)-a_(m-2)+dots+a_(n+1)-a_(n)|],[ <= |a_(m)-a_(m-1)|+|a_(m-1)-a_(m-2)|+dots+|a_(n+1)-a_(n)|],[ <= alpha^(m-1)|a_(2)-a_(1)|+alpha^(m-2)|a_(2)-a_(1)|+dots+alpha ^(n)|a_(2)-a_(1)|],[=alpha ^(n)|a_(2)-a_(1)|(1+alpha+alpha^(2)+dots+alpha^(m-n-1))],[=alpha ^(n)|a_(2)-a_(1)|(1-alpha^(m-n))/(1-alpha)],[ < alpha ^(n)|a_(2)-a_(1)|(1)/(1-alpha)]:}\begin{aligned} |a_m – a_n| &= |a_m – a_{m-1} + a_{m-1} – a_{m-2} + \ldots + a_{n+1} – a_n| \\ &\leq |a_m – a_{m-1}| + |a_{m-1} – a_{m-2}| + \ldots + |a_{n+1} – a_n| \\ &\leq \alpha^{m-1}|a_2 – a_1| + \alpha^{m-2}|a_2 – a_1| + \ldots + \alpha^n|a_2 – a_1| \\ &= \alpha^n|a_2 – a_1| (1 + \alpha + \alpha^2 + \ldots + \alpha^{m-n-1}) \\ &= \alpha^n|a_2 – a_1| \frac{1 – \alpha^{m-n}}{1 – \alpha} \\ &< \alpha^n|a_2 – a_1| \frac{1}{1 – \alpha} \end{aligned}|am−an|=|am−am−1+am−1−am−2+…+an+1−an|≤|am−am−1|+|am−1−am−2|+…+|an+1−an|≤αm−1|a2−a1|+αm−2|a2−a1|+…+αn|a2−a1|=αn|a2−a1|(1+α+α2+…+αm−n−1)=αn|a2−a1|1−αm−n1−α<αn|a2−a1|11−α
We can make this less than ϵ ϵ epsilon\epsilonϵ by choosing N N NNN large enough so that α N | a 2 − a 1 | 1 1 − α < ϵ α N | a 2 − a 1 | 1 1 − α < ϵ alpha ^(N)|a_(2)-a_(1)|(1)/(1-alpha) < epsilon\alpha^N|a_2 – a_1| \frac{1}{1 – \alpha} < \epsilonαN|a2−a1|11−α<ϵ.

Conclusion

We have shown that for any ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists an N N NNN such that for all m , n > N m , n > N m,n > Nm, n > Nm,n>N, | a m − a n | < ϵ | a m − a n | < ϵ |a_(m)-a_(n)| < epsilon|a_m – a_n| < \epsilon|am−an|<ϵ. Therefore, the sequence ( a n ) a n (a_(n))\left(a_n\right)(an) is a Cauchy sequence, as required.
1.(d) समाकल ∫ C ( z 2 + 3 z ) d z ∫ C   z 2 + 3 z d z int _(C)(z^(2)+3z)dz\int_C\left(z^2+3 z\right) d z∫C(z2+3z)dz का, ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) से ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) तक वक्र C C CCC के वामावर्त अनुगत जहाँ पर C C CCC वृत्त | z | = 2 | z | = 2 |z|=2|z|=2|z|=2 है, मान निकालिए ।
Evaluate the integral ∫ C ( z 2 + 3 z ) d z ∫ C   z 2 + 3 z d z int _(C)(z^(2)+3z)dz\int_C\left(z^2+3 z\right) d z∫C(z2+3z)dz counterclockwise from ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) to ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) along the curve C C CCC, where C C CCC is the circle | z | = 2 | z | = 2 |z|=2|z|=2|z|=2.
Answer:

Introduction

The problem asks us to evaluate the integral ∫ C ( z 2 + 3 z ) d z ∫ C   z 2 + 3 z d z int _(C)(z^(2)+3z)dz\int_C\left(z^2+3 z\right) d z∫C(z2+3z)dz counterclockwise from ( 2 , 0 ) ( 2 , 0 ) (2,0)(2,0)(2,0) to ( 0 , 2 ) ( 0 , 2 ) (0,2)(0,2)(0,2) along the curve C C CCC, where C C CCC is the circle | z | = 2 | z | = 2 |z|=2|z|=2|z|=2.

Given Parameters and Parametrization

Given | z | = 2 | z | = 2 |z|=2|z|=2|z|=2, i.e. z = 2 e i θ z = 2 e i θ z=2e^(i theta)quadz=2 e^{i \theta} \quadz=2eiθ and 0 ≤ θ ≤ π 2 0 ≤ θ ≤ π 2 quad0 <= theta <= (pi)/(2)\quad 0 \leq \theta \leq \frac{\pi}{2}0≤θ≤π2.

Integral Setup

∴ ∫ C ( z 2 + 3 z ) d z = ∫ | z | = 2 ( z 2 + 3 z ) d z ∴ ∫ C   z 2 + 3 z d z = ∫ | z | = 2   z 2 + 3 z d z :.int _(C)(z^(2)+3z)dz=int_(|z|=2)(z^(2)+3z)dz\therefore \int_C\left(z^2+3 z\right) d z =\int_{|z|=2}\left(z^2+3 z\right) d z∴∫C(z2+3z)dz=∫|z|=2(z2+3z)dz

Substitution and Simplification

= ∫ 0 π / 2 [ ( 2 e i θ ) 2 + 3 ( 2 e i θ ) ] 2 e i θ ⋅ i d θ = ∫ 0 π / 2   2 e i θ 2 + 3 2 e i θ 2 e i θ ⋅ i d θ =int_(0)^(pi//2)[(2e^(i theta))^(2)+3(2e^(i theta))]2e^(i theta)*id theta=\int_0^{\pi / 2}\left[\left(2 e^{i \theta}\right)^2+3\left(2 e^{i \theta}\right)\right] 2 e^{i \theta} \cdot i d \theta=∫0π/2[(2eiθ)2+3(2eiθ)]2eiθ⋅idθ

Evaluation of the Integral

= [ 8 i e 3 i θ 3 i + 12 i e 2 i θ 2 i ] 0 π 2 = 8 i e 3 i θ 3 i + 12 i e 2 i θ 2 i 0 π 2 =[8i(e^(3i theta))/(3i)+12 i(e^(2i theta))/(2i)]_(0)^((pi)/(2))=\left[8 i \frac{e^{3 i \theta}}{3 i}+12 i \frac{e^{2 i \theta}}{2 i}\right]_0^{\frac{\pi}{2}}=[8ie3iθ3i+12ie2iθ2i]0π2

Final Simplification

= 8 3 ( e i 3 π 2 − 1 ) + 6 ( e i π − 1 ) = 8 3 e i 3 π 2 − 1 + 6 e i π − 1 =(8)/(3)(e^(i3)pi^(2)-1)+6(e^(i pi)-1)=\frac{8}{3}\left(e^{i 3} \pi^2-1\right)+6\left(e^{i \pi}-1\right)=83(ei3π2−1)+6(eiπ−1)
= 8 3 ( − i − 1 ) + 6 ( − 2 ) = 8 3 ( − i − 1 ) + 6 ( − 2 ) =(8)/(3)(-i-1)+6(-2)=\frac{8}{3}(-i-1)+6(-2)=83(−i−1)+6(−2)
= − 44 3 − 8 3 i = − 44 3 − 8 3 i =(-44)/(3)-(8)/(3)i=\frac{-44}{3}-\frac{8}{3} i=−443−83i

Conclusion

The integral ∫ C ( z 2 + 3 z ) d z ∫ C   z 2 + 3 z d z int _(C)(z^(2)+3z)dz\int_C\left(z^2+3 z\right) d z∫C(z2+3z)dz along the curve C C CCC, where C C CCC is the circle | z | = 2 | z | = 2 |z|=2|z|=2|z|=2, is − 44 3 − 8 3 i − 44 3 − 8 3 i (-44)/(3)-(8)/(3)i\frac{-44}{3}-\frac{8}{3} i−443−83i.
1.(e) यू.पी.एस.सी. के रखरखाव विभाग ने भवन में पर्दों की आवश्यकता-पूर्ति हेतु पर्दा-कपडे के पर्याप्त संख्या में टुकड़े खरीदे हैं। प्रत्येक टुकड़े की लम्बाई 17 फुट है। पर्दों की लम्बाई के अनुसार आवश्यकता निम्नलिखित है :
पर्दे की लम्बाई (फुटों में) आवश्यक संख्या 5 700 9 400 7 300  पर्दे की लम्बाई (फुटों में) आवश्यक संख्या  5 700 9 400 7 300 {:[” पर्दे की लम्बाई (फुटों में) आवश्यक संख्या “],[5,700],[9,400],[7,300]:}\begin{array}{cc}\text { पर्दे की लम्बाई (फुटों में) आवश्यक संख्या } \\ 5 & 700 \\ 9 & 400 \\ 7 & 300\end{array} पर्दे की लम्बाई (फुटों में) आवश्यक संख्या 570094007300
टुकडों एवं सभी पर्दों की चौड़ाइयाँ समान हैं। विभिन्न रूप से काटे गये टुकड़ों की संख्या का निर्णय इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक इस प्रकार करने हेतु कि कुल कटान-हानि न्यूनतम हो, एक रैखिक प्रोग्रामन समस्या का प्रामाणिक रूप में निर्धारण कीजिए । इसका एक आधारी सुसंगत हल भी दीजिए ।
UPSC maintenance section has purchased sufficient number of curtain cloth pieces to meet the curtain requirement of its building. The length of each piece is 17 feet. The requirement according to curtain length is as follows:
Curtain length (in feet) Number required 5 700 9 400 7 300  Curtain length (in feet)   Number required  5 700 9 400 7 300 {:[” Curtain length (in feet) “,” Number required “],[5,700],[9,400],[7,300]:}\begin{array}{cc}\text { Curtain length (in feet) } & \text { Number required } \\ 5 & 700 \\ 9 & 400 \\ 7 & 300\end{array} Curtain length (in feet)  Number required 570094007300
The width of all curtains is same as that of available pieces. Form a linear programming problem in standard form that decides the number of pieces cut in different ways so that the total trim loss is minimum. Also give a basic feasible solution to it.
Answer:

Introduction

The UPSC maintenance section has purchased curtain cloth pieces, each of length 17 feet, to meet the curtain requirements of its building. The goal is to cut these 17-foot pieces into smaller lengths of 5, 9, and 7 feet to meet the specific requirements while minimizing the total trim loss. We will formulate this as a linear programming problem (LPP) in standard form.

Variables

Let x 1 x 1 x_(1)x_1x1 be the number of 17-foot pieces cut into one 5-foot piece and one 12-foot piece.
Let x 2 x 2 x_(2)x_2x2 be the number of 17-foot pieces cut into one 9-foot piece and one 8-foot piece.
Let x 3 x 3 x_(3)x_3x3 be the number of 17-foot pieces cut into one 7-foot piece and one 10-foot piece.

Objective Function

The objective is to minimize the total trim loss, which is the sum of the remaining lengths after cutting the 17-foot pieces. The total trim loss T T TTT can be represented as:
Minimize T = 12 x 1 + 8 x 2 + 10 x 3 Minimize  T = 12 x 1 + 8 x 2 + 10 x 3 “Minimize “T=12x_(1)+8x_(2)+10x_(3)\text{Minimize } T = 12x_1 + 8x_2 + 10x_3Minimize T=12x1+8x2+10x3

Constraints

  1. The number of 5-foot pieces should be at least 700:
    x 1 ≥ 700 x 1 ≥ 700 x_(1) >= 700x_1 \geq 700x1≥700
  2. The number of 9-foot pieces should be at least 400:
    x 2 ≥ 400 x 2 ≥ 400 x_(2) >= 400x_2 \geq 400x2≥400
  3. The number of 7-foot pieces should be at least 300:
    x 3 ≥ 300 x 3 ≥ 300 x_(3) >= 300x_3 \geq 300x3≥300
  4. All variables must be non-negative:
    x 1 , x 2 , x 3 ≥ 0 x 1 , x 2 , x 3 ≥ 0 x_(1),x_(2),x_(3) >= 0x_1, x_2, x_3 \geq 0x1,x2,x3≥0

Linear Programming Problem in Standard Form

Minimize T = 12 x 1 + 8 x 2 + 10 x 3 Minimize  T = 12 x 1 + 8 x 2 + 10 x 3 “Minimize “T=12x_(1)+8x_(2)+10x_(3)\text{Minimize } T = 12x_1 + 8x_2 + 10x_3Minimize T=12x1+8x2+10x3
subject to:
x 1 ≥ 700 x 2 ≥ 400 x 3 ≥ 300 x 1 , x 2 , x 3 ≥ 0 x 1 ≥ 700 x 2 ≥ 400 x 3 ≥ 300 x 1 , x 2 , x 3 ≥ 0 {:[x_(1) >= 700],[x_(2) >= 400],[x_(3) >= 300],[x_(1)”,”x_(2)”,”x_(3) >= 0]:}\begin{aligned} x_1 & \geq 700 \\ x_2 & \geq 400 \\ x_3 & \geq 300 \\ x_1, x_2, x_3 & \geq 0 \end{aligned}x1≥700x2≥400x3≥300x1,x2,x3≥0

Basic Feasible Solution

A basic feasible solution can be obtained by setting the slack variables to zero and solving the constraints for x 1 , x 2 , x 3 , x 1 , x 2 , x 3 , x_(1),x_(2),x_(3),x_1, x_2, x_3,x1,x2,x3, and T T TTT.
  • x 1 = 700 x 1 = 700 x_(1)=700x_1 = 700x1=700
  • x 2 = 400 x 2 = 400 x_(2)=400x_2 = 400x2=400
  • x 3 = 300 x 3 = 300 x_(3)=300x_3 = 300x3=300
  • T = 12 × 700 + 8 × 400 + 10 × 300 T = 12 × 700 + 8 × 400 + 10 × 300 T=12 xx700+8xx400+10 xx300T = 12 \times 700 + 8 \times 400 + 10 \times 300T=12×700+8×400+10×300
After Calculating, we get:
T = 8400 + 3200 + 3000 = 14600 T = 8400 + 3200 + 3000 = 14600 T=8400+3200+3000=14600T = 8400 + 3200 + 3000 = 14600T=8400+3200+3000=14600

Conclusion

The linear programming problem to minimize the total trim loss while meeting the curtain requirements is formulated as above. A basic feasible solution suggests cutting 700 pieces into 5-foot lengths, 400 pieces into 9-foot lengths, and 300 pieces into 7-foot lengths, resulting in a total trim loss of 14,600 feet.
2.(a) मान लीजिए G , n G , n G,nG, nG,n समूहांक का परिमित चक्रीय समूह है। तब सिद्ध कीजिए कि G G GGG के ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) जनक हैं (जहाँ पर ϕ ϕ phi\phiϕ ऑयलर ϕ ϕ phi\phiϕ-फलन है) ।
Let G G GGG be a finite cyclic group of order n n nnn. Then prove that G G GGG has ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) generators (where ϕ ϕ phi\phiϕ is Euler’s ϕ ϕ phi\phiϕ-function).
Answer:

Introduction

The problem asks us to prove that a finite cyclic group G G GGG of order n n nnn has ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) generators, where ϕ ϕ phi\phiϕ is Euler’s ϕ ϕ phi\phiϕ-function. Euler’s ϕ ϕ phi\phiϕ-function ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) is defined as the number of positive integers less than n n nnn that are relatively prime to n n nnn.

Preliminaries

Let G G GGG be a finite cyclic group of order n n nnn generated by a a aaa. That is, G = { a 0 , a 1 , a 2 , … , a n − 1 } G = { a 0 , a 1 , a 2 , … , a n − 1 } G={a^(0),a^(1),a^(2),dots,a^(n-1)}G = \{ a^0, a^1, a^2, \ldots, a^{n-1} \}G={a0,a1,a2,…,an−1}.

Generators of G G GGG

A generator g g ggg of G G GGG is an element such that every element in G G GGG can be written as a power of g g ggg. In other words, G = { g 0 , g 1 , g 2 , … , g n − 1 } G = { g 0 , g 1 , g 2 , … , g n − 1 } G={g^(0),g^(1),g^(2),dots,g^(n-1)}G = \{ g^0, g^1, g^2, \ldots, g^{n-1} \}G={g0,g1,g2,…,gn−1}.

Euler’s ϕ ϕ phi\phiϕ-Function

Euler’s ϕ ϕ phi\phiϕ-function ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) counts the number of positive integers less than n n nnn that are relatively prime to n n nnn.

Proof

  1. Claim: An element a k a k a^(k)a^kak generates G G GGG if and only if gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1.
    Proof of Claim:
    • Forward Direction: Suppose a k a k a^(k)a^kak generates G G GGG. Then ⟨ a k ⟩ = G ⟨ a k ⟩ = G (:a^(k):)=G\langle a^k \rangle = G⟨ak⟩=G, which means a k a k a^(k)a^kak has order n n nnn. By Lagrange’s theorem, the order of a k a k a^(k)a^kak must divide n n nnn. Since a k a k a^(k)a^kak has order n n nnn, k k kkk and n n nnn must be relatively prime.
    • Backward Direction: Suppose gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1. We want to show that a k a k a^(k)a^kak generates G G GGG. To do this, we need to show that every element a m a m a^(m)a^mam in G G GGG can be written as ( a k ) r ( a k ) r (a^(k))^(r)(a^k)^r(ak)r for some integer r r rrr.
    Since gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1, there exist integers p p ppp and q q qqq such that p k + q n = 1 p k + q n = 1 pk+qn=1pk + qn = 1pk+qn=1. For any a m a m a^(m)a^mam in G G GGG, we have:
    a m = a m p k = ( a k ) m p = ( a k ) r a m = a m p k = ( a k ) m p = ( a k ) r a^(m)=a^(mpk)=(a^(k))^(mp)=(a^(k))^(r)a^m = a^{mpk} = (a^k)^{mp} = (a^k)^ram=ampk=(ak)mp=(ak)r
    where r = m p r = m p r=mpr = mpr=mp. This shows that a k a k a^(k)a^kak generates G G GGG.
  2. Counting Generators: The number of generators of G G GGG is the same as the number of integers k k kkk such that 0 < k < n 0 < k < n 0 < k < n0 < k < n0<k<n and gcd ( k , n ) = 1 gcd ( k , n ) = 1 gcd(k,n)=1\gcd(k, n) = 1gcd(k,n)=1. This is precisely ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n).

Conclusion

We have proven that a finite cyclic group G G GGG of order n n nnn has ϕ ( n ) ϕ ( n ) phi(n)\phi(n)ϕ(n) generators. The proof relies on the properties of Euler’s ϕ ϕ phi\phiϕ-function and the definition of a cyclic group.
2.(b) सिद्ध कीजिए कि फलन f ( x ) = sin ⁡ x 2 f ( x ) = sin ⁡ x 2 f(x)=sin x^(2)f(x)=\sin x^2f(x)=sin⁡x2 अंतराल [ 0 , ∞ [ 0 , ∞ [0,oo[0, \infty[0,∞ [ पर एकसमान संतत नहीं है।
Prove that the function f ( x ) = sin ⁡ x 2 f ( x ) = sin ⁡ x 2 f(x)=sin x^(2)f(x)=\sin x^2f(x)=sin⁡x2 is not uniformly continuous on the interval [ 0 , ∞ [ [ 0 , ∞ [ [0,oo[[0, \infty[[0,∞[.
Answer:

Introduction

The problem asks us to prove that the function f ( x ) = sin ⁡ ( x 2 ) f ( x ) = sin ⁡ ( x 2 ) f(x)=sin(x^(2))f(x) = \sin(x^2)f(x)=sin⁡(x2) is not uniformly continuous on the interval [ 0 , ∞ ) [ 0 , ∞ ) [0,oo)[0, \infty)[0,∞).

Definition of Uniform Continuity

A function f ( x ) f ( x ) f(x)f(x)f(x) is said to be uniformly continuous on an interval I I III if for every ϵ > 0 ϵ > 0 epsilon > 0\epsilon > 0ϵ>0, there exists a δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that for all x , y x , y x,yx, yx,y in I I III, if | x − y | < δ | x − y | < δ |x-y| < delta|x – y| < \delta|x−y|<δ, then | f ( x ) − f ( y ) | < ϵ | f ( x ) − f ( y ) | < ϵ |f(x)-f(y)| < epsilon|f(x) – f(y)| < \epsilon|f(x)−f(y)|<ϵ.

Proof by Contradiction

To prove that f ( x ) = sin ⁡ ( x 2 ) f ( x ) = sin ⁡ ( x 2 ) f(x)=sin(x^(2))f(x) = \sin(x^2)f(x)=sin⁡(x2) is not uniformly continuous on [ 0 , ∞ ) [ 0 , ∞ ) [0,oo)[0, \infty)[0,∞), we’ll use a proof by contradiction. Assume that f ( x ) f ( x ) f(x)f(x)f(x) is uniformly continuous on [ 0 , ∞ ) [ 0 , ∞ ) [0,oo)[0, \infty)[0,∞).
  1. Assumption: Assume f ( x ) = sin ⁡ ( x 2 ) f ( x ) = sin ⁡ ( x 2 ) f(x)=sin(x^(2))f(x) = \sin(x^2)f(x)=sin⁡(x2) is uniformly continuous on [ 0 , ∞ ) [ 0 , ∞ ) [0,oo)[0, \infty)[0,∞).
  2. Choose ϵ ϵ epsilon\epsilonϵ: Let ϵ = 1 2 ϵ = 1 2 epsilon=(1)/(2)\epsilon = \frac{1}{2}ϵ=12.
  3. Find δ δ delta\deltaδ: According to the definition of uniform continuity, there must exist a δ > 0 δ > 0 delta > 0\delta > 0δ>0 such that for all x , y x , y x,yx, yx,y in [ 0 , ∞ ) [ 0 , ∞ ) [0,oo)[0, \infty)[0,∞), if | x − y | < δ | x − y | < δ |x-y| < delta|x – y| < \delta|x−y|<δ, then | sin ⁡ ( x 2 ) − sin ⁡ ( y 2 ) | < 1 2 | sin ⁡ ( x 2 ) − sin ⁡ ( y 2 ) | < 1 2 |sin(x^(2))-sin(y^(2))| < (1)/(2)|\sin(x^2) – \sin(y^2)| < \frac{1}{2}|sin⁡(x2)−sin⁡(y2)|<12.
  4. Construct Counterexample: Consider the sequence x n = n π x n = n π x_(n)=sqrt(n pi)x_n = \sqrt{n\pi}xn=nπ and y n = ( n + 1 ) π y n = ( n + 1 ) π y_(n)=sqrt((n+1)pi)y_n = \sqrt{(n+1)\pi}yn=(n+1)π. We have:
    | x n − y n | = | n π − ( n + 1 ) π | = | π ( n − n + 1 ) | | x n − y n | = n π − ( n + 1 ) π = π n − n + 1 |x_(n)-y_(n)|=|sqrt(n pi)-sqrt((n+1)pi)|=|sqrtpi(sqrtn-sqrt(n+1))||x_n – y_n| = \left| \sqrt{n\pi} – \sqrt{(n+1)\pi} \right| = \left| \sqrt{\pi} \left( \sqrt{n} – \sqrt{n+1} \right) \right||xn−yn|=|nπ−(n+1)π|=|π(n−n+1)|
    As n n nnn approaches infinity, | x n − y n | | x n − y n | |x_(n)-y_(n)||x_n – y_n||xn−yn| approaches 0, which means for sufficiently large n n nnn, | x n − y n | < δ | x n − y n | < δ |x_(n)-y_(n)| < delta|x_n – y_n| < \delta|xn−yn|<δ.
  5. Evaluate f ( x n ) − f ( y n ) f ( x n ) − f ( y n ) f(x_(n))-f(y_(n))f(x_n) – f(y_n)f(xn)−f(yn): We have f ( x n ) = sin ⁡ ( n π ) = 0 f ( x n ) = sin ⁡ ( n π ) = 0 f(x_(n))=sin(n pi)=0f(x_n) = \sin(n\pi) = 0f(xn)=sin⁡(nπ)=0 and f ( y n ) = sin ⁡ ( ( n + 1 ) π ) = 0 f ( y n ) = sin ⁡ ( ( n + 1 ) π ) = 0 f(y_(n))=sin((n+1)pi)=0f(y_n) = \sin((n+1)\pi) = 0f(yn)=sin⁡((n+1)π)=0. Therefore, | f ( x n ) − f ( y n ) | = 0 | f ( x n ) − f ( y n ) | = 0 |f(x_(n))-f(y_(n))|=0|f(x_n) – f(y_n)| = 0|f(xn)−f(yn)|=0.
  6. Contradiction: The function f ( x ) = sin ⁡ ( x 2 ) f ( x ) = sin ⁡ ( x 2 ) f(x)=sin(x^(2))f(x) = \sin(x^2)f(x)=sin⁡(x2) oscillates infinitely many times as x x xxx approaches infinity. This means that for any δ > 0 δ > 0 delta > 0\delta > 0δ>0, we can find points x , y x , y x,yx, yx,y such that | x − y | < δ | x − y | < δ |x-y| < delta|x – y| < \delta|x−y|<δ but | sin ⁡ ( x 2 ) − sin ⁡ ( y 2 ) | | sin ⁡ ( x 2 ) − sin ⁡ ( y 2 ) | |sin(x^(2))-sin(y^(2))||\sin(x^2) – \sin(y^2)||sin⁡(x2)−sin⁡(y2)| is close to 1, contradicting the assumption that f ( x ) f ( x ) f(x)f(x)f(x) is uniformly continuous.

Conclusion

We have shown that assuming f ( x ) = sin ⁡ ( x 2 ) f ( x ) = sin ⁡ ( x 2 ) f(x)=sin(x^(2))f(x) = \sin(x^2)f(x)=sin⁡(x2) is uniformly continuous leads to a contradiction. Therefore, f ( x ) = sin ⁡ ( x 2 ) f ( x ) = sin ⁡ ( x 2 ) f(x)=sin(x^(2))f(x) = \sin(x^2)f(x)=sin⁡(x2) is not uniformly continuous on the interval [ 0 , ∞ ) [ 0 , ∞ ) [0,oo)[0, \infty)[0,∞).
2.(c) कन्टूर समाकलन का उपयोग कर, समाकल ∫ 0 2 π 1 3 + 2 sin ⁡ θ d θ ∫ 0 2 π   1 3 + 2 sin ⁡ θ d θ int_(0)^(2pi)(1)/(3+2sin theta)d theta\int_0^{2 \pi} \frac{1}{3+2 \sin \theta} d \theta∫02π13+2sin⁡θdθ का मान ज्ञात कीजिए ।
Using contour integration, evaluate the integral ∫ 0 2 π 1 3 + 2 sin ⁡ θ d θ ∫ 0 2 π   1 3 + 2 sin ⁡ θ d θ int_(0)^(2pi)(1)/(3+2sin theta)d theta\int_0^{2 \pi} \frac{1}{3+2 \sin \theta} d \theta∫02π13+2sin⁡θdθ.
Answer:
Introduction:
Using contour integration, evaluate the integral
∫ 0 2 π 1 3 + 2 sin ⁡ θ d θ . ∫ 0 2 π   1 3 + 2 sin ⁡ θ d θ . int_(0)^(2pi)(1)/(3+2sin theta)d theta.\int_0^{2 \pi} \frac{1}{3+2 \sin \theta} d \theta.∫02π13+2sin⁡θdθ.
Answer:
I = ∫ 0 2 π d θ 3 + 2 sin ⁡ θ I = ∫ 0 2 π   d θ 3 + 2 sin ⁡ θ I=int_(0)^(2pi)(dtheta)/(3+2sin theta)I=\int_0^{2 \pi} \frac{\mathbf{d} \theta}{3+2 \sin \theta}I=∫02πdθ3+2sin⁡θ
Step 1: Change of Variable
Consider z = e i θ ⇒ d z = i e i θ d θ z = e i θ ⇒ d z = i e i θ d θ z=e^(i theta)=>dz=ie^(i theta)d thetaz=e^{i \theta} \Rightarrow \mathbf{d} z=i e^{i \theta} d \thetaz=eiθ⇒dz=ieiθdθ.
⇒ d θ = d z i z , sin ⁡ θ = z − z − 1 2 i = z 2 − 1 2 i z ⇒ d θ = d z i z , sin ⁡ θ = z − z − 1 2 i = z 2 − 1 2 i z =>d theta=(dz)/(iz),sin theta=(z-z^(-1))/(2i)=(z^(2)-1)/(2iz)\Rightarrow d \theta=\frac{\mathbf{d} z}{i z}, \sin \theta=\frac{z-z^{-1}}{2 i}=\frac{z^2-1}{2 i z}⇒dθ=dziz,sin⁡θ=z−z−12i=z2−12iz
Step 2: Substitution
Then
I = ∫ 1 3 + 2 ( z 2 − 1 ) 2 i z d z i z = ∫ 1 3 i z + z 2 − 1 i z d z i z = ∫ c d z z 2 + 3 i z − 1 = ∫ c f ( z ) d z I = ∫ 1 3 + 2 z 2 − 1 2 i z d z i z = ∫ 1 3 i z + z 2 − 1 i z d z i z = ∫ c   d z z 2 + 3 i z − 1 = ∫ c   f ( z ) d z I=int(1)/(3+(2(z^(2)-1))/(2iz))(dz)/(iz)=int((1)/(3iz+z^(2)-1))/(iz)(dz)/(iz)=int _(c)(dz)/(z^(2)+3iz-1)=int _(c)f(z)dzI=\int \frac{1}{3+\frac{2\left(z^2-1\right)}{2 i z}} \frac{\mathbf{d} z}{i z}=\int \frac{\frac{1}{3 i z+z^2-1}}{i z} \frac{\mathbf{d} z}{i z}=\int_c \frac{\mathbf{d} z}{z^2+3 i z-1}=\int_c f(z) \mathbf{d} zI=∫13+2(z2−1)2izdziz=∫13iz+z2−1izdziz=∫cdzz2+3iz−1=∫cf(z)dz
Step 3: Finding Poles
f ( z ) f ( z ) f(z)f(z)f(z) has poles when z 2 + 3 i z − 1 = 0 z 2 + 3 i z − 1 = 0 z^(2)+3iz-1=0z^2+3 i z-1=0z2+3iz−1=0
z = − 3 i ± 9 i 2 − 4 ( − 1 ) ( 1 ) 2 z = − 3 i ± − 9 + 4 2 = − 3 i ± 5 i 2 z = − 3 i ± 9 i 2 − 4 ( − 1 ) ( 1 ) 2 z = − 3 i ± − 9 + 4 2 = − 3 i ± 5 i 2 {:[z=(-3i+-sqrt(9i^(2)-4(-1)(1)))/(2)],[z=(-3i+-sqrt(-9+4))/(2)=(-3i+-sqrt5i)/(2)]:}\begin{aligned} & z=\frac{-3 i \pm \sqrt{9 i^2-4(-1)(1)}}{2} \\ & z=\frac{-3 i \pm \sqrt{-9+4}}{2}=\frac{-3 i \pm \sqrt{5} i}{2} \end{aligned}z=−3i±9i2−4(−1)(1)2z=−3i±−9+42=−3i±5i2
Thus z 1 = ( − 3 + 5 ) i 2 , z 2 = ( − 3 − 5 ) i 2 z 1 = ( − 3 + 5 ) i 2 , z 2 = ( − 3 − 5 ) i 2 z_(1)=((-3+sqrt5)i)/(2),z_(2)=((-3-sqrt5)i)/(2)z_1=\frac{(-3+\sqrt{5}) i}{2}, z_2=\frac{(-3-\sqrt{5}) i}{2}z1=(−3+5)i2,z2=(−3−5)i2 are poles.
Step 4: Residue Calculation
But only z 1 z 1 z_(1)z_1z1 lies inside C C CCC.
The residue at z = z 1 z = z 1 z=z_(1)z=z_1z=z1
lim z → z 1 ( z − z 1 ) 1 ( z − z 1 ) ( z − z 2 ) lim z → z 1 1 z − z 2 = 1 z 1 − z 2 = 1 5 i lim z → z 1   z − z 1 1 z − z 1 z − z 2 lim z → z 1   1 z − z 2 = 1 z 1 − z 2 = 1 5 i {:[lim_(z rarrz_(1))(z-z_(1))(1)/((z-z_(1))(z-z_(2)))],[lim_(z rarrz_(1))(1)/(z-z_(2))=(1)/(z_(1)-z_(2))=(1)/(sqrt5i)]:}\begin{aligned} & \lim _{z \rightarrow z_1}\left(z-z_1\right) \frac{1}{\left(z-z_1\right)\left(z-z_2\right)} \\ & \lim _{z \rightarrow z_1} \frac{1}{z-z_2}=\frac{1}{z_1-z_2}=\frac{1}{\sqrt{5}i} \end{aligned}limz→z1(z−z1)1(z−z1)(z−z2)limz→z11z−z2=1z1−z2=15i
Step 5: Applying Residue Theorem
By residue theorem
∫ c f ( z ) d z = 2 π i × 1 5 i = 2 π 5 ∫ c   f ( z ) d z = 2 π i × 1 5 i = 2 π 5 int _(c)f(z)dz=2pi i xx(1)/(sqrt5i)=(2pi)/(sqrt5)\int_c f(z) \mathbf{d} z=2 \pi i \times \frac{1}{\sqrt{5}i}=\frac{2 \pi}{\sqrt{5}}∫cf(z)dz=2πi×15i=2π5
Conclusion:
Therefore,
∫ 0 2 π d θ 3 + 2 sin ⁡ θ = 2 π 5 ∫ 0 2 π   d θ 3 + 2 sin ⁡ θ = 2 π 5 int_(0)^(2pi)(dtheta)/(3+2sin theta)=(2pi)/(sqrt5)\int_0^{2 \pi} \frac{\mathbf{d} \theta}{3+2 \sin \theta}=\frac{2 \pi}{\sqrt{5}}∫02πdθ3+2sin⁡θ=2π5
  1. (a) मान लीजिए R , p ( > 0 ) R , p ( > 0 ) R,p( > 0)R, p(>0)R,p(>0) अभिलक्षण का एक परिमित क्षेत्र है । दर्शाइए कि f ( a ) = a p , ∀ a ∈ R f ( a ) = a p , ∀ a ∈ R f(a)=a^(p),AA a in Rf(a)=a^p, \forall a \in Rf(a)=ap,∀a∈R द्वारा परिभाषित प्रतिचित्रण f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R एकैक समाकारी है ।
Let R R RRR be a finite field of characteristic p ( > 0 ) p ( > 0 ) p( > 0)p(>0)p(>0). Show that the mapping f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R defined by f ( a ) = a p , ∀ a ∈ R f ( a ) = a p , ∀ a ∈ R f(a)=a^(p),AA a in Rf(a)=a^p, \forall a \in Rf(a)=ap,∀a∈R is an isomorphism.
Answer:

Introduction

The problem asks us to prove that the mapping f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R defined by f ( a ) = a p f ( a ) = a p f(a)=a^(p)f(a) = a^pf(a)=ap for all a ∈ R a ∈ R a in Ra \in Ra∈R is an isomorphism, where R R RRR is a finite field of characteristic p > 0 p > 0 p > 0p > 0p>0.

Definitions

  1. Field: A set R R RRR with two operations + + +++ and × × xx\times× that satisfy the field axioms.
  2. Characteristic: A field R R RRR has characteristic p p ppp if p p ppp is the smallest positive integer such that p ⋅ a = 0 p ⋅ a = 0 p*a=0p \cdot a = 0p⋅a=0 for all a ∈ R a ∈ R a in Ra \in Ra∈R. If no such p p ppp exists, the characteristic is zero.
  3. Isomorphism: A bijective map f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R that preserves the field operations.

Properties Needed

  1. Frobenius Endomorphism: In a field of characteristic p p ppp, ( a + b ) p = a p + b p ( a + b ) p = a p + b p (a+b)^(p)=a^(p)+b^(p)(a+b)^p = a^p + b^p(a+b)p=ap+bp and ( a b ) p = a p b p ( a b ) p = a p b p (ab)^(p)=a^(p)b^(p)(ab)^p = a^p b^p(ab)p=apbp.

Proof

To prove that f f fff is an isomorphism, we need to show that f f fff is a bijective map that preserves addition and multiplication.

1. f f fff is a Homomorphism

  1. Preservation of Addition:
    f ( a + b ) = ( a + b ) p = a p + b p = f ( a ) + f ( b ) f ( a + b ) = ( a + b ) p = a p + b p = f ( a ) + f ( b ) f(a+b)=(a+b)^(p)=a^(p)+b^(p)=f(a)+f(b)f(a+b) = (a+b)^p = a^p + b^p = f(a) + f(b)f(a+b)=(a+b)p=ap+bp=f(a)+f(b)
  2. Preservation of Multiplication:
    f ( a b ) = ( a b ) p = a p b p = f ( a ) f ( b ) f ( a b ) = ( a b ) p = a p b p = f ( a ) f ( b ) f(ab)=(ab)^(p)=a^(p)b^(p)=f(a)f(b)f(ab) = (ab)^p = a^p b^p = f(a) f(b)f(ab)=(ab)p=apbp=f(a)f(b)

2. f f fff is Injective (One-to-One)

Suppose f ( a ) = f ( b ) f ( a ) = f ( b ) f(a)=f(b)f(a) = f(b)f(a)=f(b). Then a p = b p a p = b p a^(p)=b^(p)a^p = b^pap=bp which implies a p − b p = 0 a p − b p = 0 a^(p)-b^(p)=0a^p – b^p = 0ap−bp=0. Since R R RRR is a field, it has no zero divisors, and we can factor a p − b p a p − b p a^(p)-b^(p)a^p – b^pap−bp as ( a − b ) p ( a − b ) p (a-b)^(p)(a-b)^p(a−b)p. This means a − b = 0 a − b = 0 a-b=0a-b = 0a−b=0 or a = b a = b a=ba = ba=b, proving that f f fff is injective.

3. f f fff is Surjective (Onto)

Since R R RRR is finite and f f fff is injective, f f fff must also be surjective. Alternatively, for any b ∈ R b ∈ R b in Rb \in Rb∈R, b = b p 2 b = b p 2 b=b^(p^(2))b = b^{p^2}b=bp2 (by Fermat’s Little Theorem or the fact that R ∗ R ∗ R^(**)R^*R∗, the multiplicative group of R R RRR, has order p n − 1 p n − 1 p^(n)-1p^n – 1pn−1). Thus, f ( b p − 1 ) = ( b p − 1 ) p = b p = b f ( b p − 1 ) = ( b p − 1 ) p = b p = b f(b^(p-1))=(b^(p-1))^(p)=b^(p)=bf(b^{p-1}) = (b^{p-1})^p = b^p = bf(bp−1)=(bp−1)p=bp=b, showing that f f fff is surjective.

Conclusion

We have shown that f f fff preserves both addition and multiplication, and is both injective and surjective. Therefore, f : R → R f : R → R f:R rarr Rf: R \rightarrow Rf:R→R defined by f ( a ) = a p f ( a ) = a p f(a)=a^(p)f(a) = a^pf(a)=ap is an isomorphism.
3.(b) एकधा विधि के द्वारा निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
न्यूनतमीकरण कीजिए z = − 6 X 1 − 2 X 2 − 5 X 3 z = − 6 X 1 − 2 X 2 − 5 X 3 z=-6X_(1)-2X_(2)-5X_(3)z=-6 X_1-2 X_2-5 X_3z=−6X1−2X2−5X3
बशर्ते कि
2 X 1 − 3 X 2 + X 3 ⩽ 14 − 4 X 1 + 4 X 2 + 10 X 3 ⩽ 46 2 X 1 + 2 X 2 − 4 X 3 ⩽ 37 X 1 ⩾ 2 , X 2 ⩾ 1 , X 3 ⩾ 3 2 X 1 − 3 X 2 + X 3 ⩽ 14 − 4 X 1 + 4 X 2 + 10 X 3 ⩽ 46 2 X 1 + 2 X 2 − 4 X 3 ⩽ 37 X 1 ⩾ 2 , X 2 ⩾ 1 , X 3 ⩾ 3 {:[2X_(1)-3X_(2)+X_(3) <= 14],[-4X_(1)+4X_(2)+10X_(3) <= 46],[2X_(1)+2X_(2)-4X_(3) <= 37],[X_(1) >= 2″,”X_(2) >= 1″,”X_(3) >= 3]:}\begin{gathered} 2 X_1-3 X_2+X_3 \leqslant 14\\ -4 X_1+4 X_2+10 X_3 \leqslant 46 \\ 2 X_1+2 X_2-4 X_3 \leqslant 37 \\ X_1 \geqslant 2, X_2 \geqslant 1, X_3 \geqslant 3 \end{gathered}2X1−3X2+X3⩽14−4X1+4X2+10X3⩽462X1+2X2−4X3⩽37X1⩾2,X2⩾1,X3⩾3
Solve the linear programming problem using simplex method:
Minimize z = − 6 X 1 − 2 X 2 − 5 X 3 z = − 6 X 1 − 2 X 2 − 5 X 3 z=-6X_(1)-2X_(2)-5X_(3)z=-6 X_1-2 X_2-5 X_3z=−6X1−2X2−5X3
subject to 2 X 1 − 3 X 2 + X 3 ⩽ 14 2 X 1 − 3 X 2 + X 3 ⩽ 14 quad2X_(1)-3X_(2)+X_(3) <= 14\quad 2 X_1-3 X_2+X_3 \leqslant 142X1−3X2+X3⩽14
− 4 X 1 + 4 X 2 + 10 X 3 ⩽ 46 2 X 1 + 2 X 2 − 4 X 3 ⩽ 37 X 1 ⩾ 2 , X 2 ⩾ 1 , X 3 ⩾ 3 − 4 X 1 + 4 X 2 + 10 X 3 ⩽ 46 2 X 1 + 2 X 2 − 4 X 3 ⩽ 37 X 1 ⩾ 2 , X 2 ⩾ 1 , X 3 ⩾ 3 {:[-4X_(1)+4X_(2)+10X_(3) <= 46],[2X_(1)+2X_(2)-4X_(3) <= 37],[X_(1) >= 2″,”X_(2) >= 1″,”X_(3) >= 3]:}\begin{gathered} -4 X_1+4 X_2+10 X_3 \leqslant 46 \\ 2 X_1+2 X_2-4 X_3 \leqslant 37 \\ X_1 \geqslant 2, X_2 \geqslant 1, X_3 \geqslant 3 \end{gathered}−4X1+4X2+10X3⩽462X1+2X2−4X3⩽37X1⩾2,X2⩾1,X3⩾3
Answer:

Introduction

The given Linear Programming Problem (LPP) aims to minimize the objective function z = − 6 X 1 − 2 X 2 − 5 X 3 z = − 6 X 1 − 2 X 2 − 5 X 3 z=-6X_(1)-2X_(2)-5X_(3)z = -6X_1 – 2X_2 – 5X_3z=−6X1−2X2−5X3 subject to certain constraints and variable bounds. The problem is transformed to a standard form by introducing slack variables and then solved using the Simplex method.

Problem Transformation

Let x 1 + 2 = X 1 , x 2 + 1 = X 2 x 1 + 2 = X 1 , x 2 + 1 = X 2 x_(1)+2=X_(1),x_(2)+1=X_(2)x_1+2=X_1, x_2+1=X_2x1+2=X1,x2+1=X2 and x 3 + 1 = X 3 x 3 + 1 = X 3 x_(3)+1=X_(3)x_3+1=X_3x3+1=X3
Then, the objective function becomes:
min Z ∗ = − 6 ( x 1 + 2 ) − 2 ( x 2 + 1 ) − 5 ( x 3 + 3 ) = − 6 x 1 − 12 − 12 x 2 − 2 − 5 x 3 − 15 = − 6 x 1 − 2 x 2 − 5 x 3 − 29 min Z ∗ = − 6 x 1 + 2 − 2 x 2 + 1 − 5 x 3 + 3 = − 6 x 1 − 12 − 12 x 2 − 2 − 5 x 3 − 15 = − 6 x 1 − 2 x 2 − 5 x 3 − 29 {:[ minZ^(**)=-6(x_(1)+2)-2(x_(2)+1)-5(x_(3)+3)],[=-6x_(1)-12-12x_(2)-2-5x_(3)-15],[=-6x_(1)-2x_(2)-5x_(3)-29]:}\begin{aligned} & \min Z^*=-6\left(x_1+2\right)-2\left(x_2+1\right)-5\left(x_3+3\right) \\ & =-6 x_1-12-12 x_2-2-5 x_3-15 \\ & =-6 x_1-2 x_2-5 x_3-29 \end{aligned}minZ∗=−6(x1+2)−2(x2+1)−5(x3+3)=−6x1−12−12x2−2−5x3−15=−6x1−2x2−5x3−29
Subject to the transformed constraints:
2 ( x 1 + 2 ) − 3 ( x 2 + 1 ) + ( x 3 + 3 ) ⩽ 14 ⇒ 2 x 1 − 3 x 2 + x 3 ⩽ 10 2 x 1 + 2 − 3 x 2 + 1 + x 3 + 3 ⩽ 14 ⇒ 2 x 1 − 3 x 2 + x 3 ⩽ 10 2(x_(1)+2)-3(x_(2)+1)+(x_(3)+3) <= 14=>2x_(1)-3x_(2)+x_(3) <= 102\left(x_1+2\right)-3\left(x_2+1\right)+\left(x_3+3\right) \leqslant 14 \Rightarrow 2 x_1-3 x_2+x_3 \leqslant 102(x1+2)−3(x2+1)+(x3+3)⩽14⇒2x1−3x2+x3⩽10
− 4 ( x 1 + 2 ) + 4 ( x 2 + 1 ) + 10 ( x 3 + 3 ) ⩽ 46 ⇒ − 4 x 1 + 4 x 2 + 10 x 3 ⩽ 20 − 4 x 1 + 2 + 4 x 2 + 1 + 10 x 3 + 3 ⩽ 46 ⇒ − 4 x 1 + 4 x 2 + 10 x 3 ⩽ 20 -4(x_(1)+2)+4(x_(2)+1)+10(x_(3)+3) <= 46=>-4x_(1)+4x_(2)+10x_(3) <= 20-4\left(x_1+2\right)+4\left(x_2+1\right)+10\left(x_3+3\right) \leqslant 46 \Rightarrow -4 x_1+4 x_2+10 x_3 \leqslant 20−4(x1+2)+4(x2+1)+10(x3+3)⩽46⇒−4x1+4x2+10x3⩽20
2 ( x 1 + 2 ) + 2 ( x 2 + 1 ) − 4 ( x 3 + 3 ) ⩽ 37 ⇒ 2 x 1 + 2 x 2 − 4 x 3 ⩽ 43 2 x 1 + 2 + 2 x 2 + 1 − 4 x 3 + 3 ⩽ 37 ⇒ 2 x 1 + 2 x 2 − 4 x 3 ⩽ 43 2(x_(1)+2)+2(x_(2)+1)-4(x_(3)+3) <= 37=>2x_(1)+2x_(2)-4x_(3) <= 432\left(x_1+2\right)+2\left(x_2+1\right)-4\left(x_3+3\right) \leqslant 37 \Rightarrow 2 x_1+2 x_2-4 x_3 \leqslant 432(x1+2)+2(x2+1)−4(x3+3)⩽37⇒2x1+2x2−4x3⩽43
x 1 , x 2 , x 3 ⩾ 0 x 1 , x 2 , x 3 ⩾ 0 x_(1),x_(2),x_(3) >= 0x_1, x_2, x_3 \geqslant 0x1,x2,x3⩾0
After introducing slack variables
Min Z ∗ = − 6 x 1 − 2 x 2 − 5 x 3 + 0 S 1 + 0 S 2 + 0 S 3 subject to 2 x 1 − 3 x 2 + x 3 + S 1 = 10 − 4 x 1 + 4 x 2 + 10 x 3 + S 2 = 20 2 x 1 + 2 x 2 − 4 x 3 + S 3 = 43 and x 1 , x 2 , x 3 , S 1 , S 2 , S 3 ≥ 0 Min Z ∗ = − 6 x 1 − 2 x 2 − 5 x 3 + 0 S 1 + 0 S 2 + 0 S 3  subject to  2 x 1 − 3 x 2 + x 3 + S 1 = 10 − 4 x 1 + 4 x 2 + 10 x 3 + S 2 = 20 2 x 1 + 2 x 2 − 4 x 3 + S 3 = 43  and  x 1 , x 2 , x 3 , S 1 , S 2 , S 3 ≥ 0 {:[MinZ^(**)=-6x_(1)-2x_(2)-5x_(3)+0S_(1)+0S_(2)+0S_(3)],[” subject to “],[qquad{:[2x_(1)-3x_(2)+x_(3)+S_(1)=10],[-4x_(1)+4x_(2)+10x_(3)+S_(2)=20],[2x_(1)+2x_(2)-4x_(3)+S_(3)=43]:}],[” and “x_(1)”,”x_(2)”,”x_(3)”,”S_(1)”,”S_(2)”,”S_(3) >= 0]:}\begin{aligned} & \operatorname{Min} Z^*=-6 x_1-2 x_2-5 x_3+0 S_1+0 S_2+0 S_3 \\ & \text { subject to } \\ & \qquad \begin{aligned} 2 x_1-3 x_2+x_3+S_1 & =10 \\ -4 x_1+4 x_2+10 x_3+S_2 & =20 \\ 2 x_1+2 x_2-4 x_3 & +S_3=43 \end{aligned} \\ & \text { and } x_1, x_2, x_3, S_1, S_2, S_3 \geq 0 \end{aligned}MinZ∗=−6x1−2x2−5x3+0S1+0S2+0S3 subject to 2x1−3x2+x3+S1=10−4x1+4x2+10x3+S2=202x1+2x2−4x3+S3=43 and x1,x2,x3,S1,S2,S3≥0
Iteration-1 C j C j C_(j)C_jCj-6-2-5000
B B BBB C B C B C_(B)C_BCB X B X B X_(B)X_BXB x 1 x 1 x_(1)x_1x1 x 2 x 2 x_(2)x_2x2 x 3 x 3 x_(3)x_3x3 S 1 S 1 S_(1)S_1S1 S 2 S 2 S_(2)S_2S2 S 3 S 3 S_(3)S_3S3 MinRatio X B x 1  MinRatio  X B x 1 {:[” MinRatio “],[(X_(B))/(x_(1))]:}\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_1}\end{array} MinRatio XBx1
S 1 S 1 S_(1)S_1S1010(2)-31100 10 2 = 5 → 10 2 = 5 → (10)/(2)=5rarr\frac{10}{2}=5 \rightarrow102=5→
S 2 S 2 S_(2)S_2S2020-4410010—
S 3 S 3 S_(3)S_3S304322-4001 43 2 = 21.5 43 2 = 21.5 (43)/(2)=21.5\frac{43}{2}=21.5432=21.5
Z = 0 Z = 0 Z=0Z=0Z=0 Z j Z j Z_(j)Z_jZj00 0 0 0\mathbf{0}0000
C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj − 6 ↑ − 6 ↑ -6uarr-6 \uparrow−6↑-2-5000
Iteration-1 C_(j) -6 -2 -5 0 0 0 B C_(B) X_(B) x_(1) x_(2) x_(3) S_(1) S_(2) S_(3) ” MinRatio (X_(B))/(x_(1))” S_(1) 0 10 (2) -3 1 1 0 0 (10)/(2)=5rarr S_(2) 0 20 -4 4 10 0 1 0 — S_(3) 0 43 2 2 -4 0 0 1 (43)/(2)=21.5 Z=0 Z_(j) 0 0 0 0 0 0 C_(j)-Z_(j) -6uarr -2 -5 0 0 0 | Iteration-1 | | $C_j$ | -6 | -2 | -5 | 0 | 0 | 0 | | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $B$ | $C_B$ | $X_B$ | $x_1$ | $x_2$ | $x_3$ | $S_1$ | $S_2$ | $S_3$ | $\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_1}\end{array}$ | | $S_1$ | 0 | 10 | (2) | -3 | 1 | 1 | 0 | 0 | $\frac{10}{2}=5 \rightarrow$ | | $S_2$ | 0 | 20 | -4 | 4 | 10 | 0 | 1 | 0 | — | | $S_3$ | 0 | 43 | 2 | 2 | -4 | 0 | 0 | 1 | $\frac{43}{2}=21.5$ | | $Z=0$ | | $Z_j$ | 0 | 0 | $\mathbf{0}$ | 0 | 0 | 0 | | | | | $C_j-Z_j$ | $-6 \uparrow$ | -2 | -5 | 0 | 0 | 0 | |
Negative minimum C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj is -6 and its column index is 1 . So, the entering variable is x 1 x 1 x_(1)x_1x1.
Minimum ratio is 5 and its row index is 1 . So, the leaving basis variable is S 1 S 1 S_(1)S_1S1.
∴ ∴ :.\therefore∴ The pivot element is 2 .
Entering = x 1 = x 1 =x_(1)=x_1=x1, Departing = S 1 = S 1 =S_(1)=S_1=S1, Key Element = 2 = 2 =2=2=2
R 1 ( new ) = R 1 ( old ) ÷ 2 R 2 ( new ) = R 2 ( old ) + 4 R 1 ( new ) R 3 (new ) = R 3 (old ) − 2 R 1 ( new ) R 1 (  new  ) = R 1 (  old  ) ÷ 2 R 2 (  new  ) = R 2 (  old  ) + 4 R 1 (  new  ) R 3  (new  = R 3  (old  − 2 R 1 (  new  ) {:[R_(1)(” new “)=R_(1)(” old “)-:2],[R_(2)(” new “)=R_(2)(” old “)+4R_(1)(” new “)],[{:R_(3)” (new “)=R_(3)” (old “)-2R_(1)(” new “)]:}\begin{aligned} & R_1(\text { new })=R_1(\text { old }) \div 2 \\ & R_2(\text { new })=R_2(\text { old })+4 R_1(\text { new }) \\ & \left.\left.R_3 \text { (new }\right)=R_3 \text { (old }\right)-2 R_1(\text { new }) \end{aligned}R1( new )=R1( old )÷2R2( new )=R2( old )+4R1( new )R3 (new )=R3 (old )−2R1( new )
Iteration-2 C j C j C_(j)C_jCj-6-2-5000
B B BBB C B C B C_(B)C_BCB X B X B X_(B)X_BXB x 1 x 1 x_(1)x_1x1 x 2 x 2 x_(2)x_2x2 x 3 x 3 x_(3)x_3x3 S 1 S 1 S_(1)S_1S1 S 2 S 2 S_(2)S_2S2 S 3 S 3 S_(3)S_3S3 MinRatio X B x 2  MinRatio  X B x 2 {:[” MinRatio “],[(X_(B))/(x_(2))]:}\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_2}\end{array} MinRatio XBx2
x 1 x 1 x_(1)x_1x1-651-1.50.50.500—
S 2 S 2 S_(2)S_2S20400-212210—
S 3 S 3 S_(3)S_3S30330(5)-5-101 33 5 = 6.6 → 33 5 = 6.6 → (33)/(5)=6.6 rarr\frac{33}{5}=6.6 \rightarrow335=6.6→
Z = − 3 0 Z = − 3 0 Z=-30Z=-\mathbf{3 0}Z=−30 Z j Z j Z_(j)Z_jZj-69-3-300
C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj0 − 11 ↑ − 11 ↑ -11 uarr-11 \uparrow−11↑-2300
Iteration-2 C_(j) -6 -2 -5 0 0 0 B C_(B) X_(B) x_(1) x_(2) x_(3) S_(1) S_(2) S_(3) ” MinRatio (X_(B))/(x_(2))” x_(1) -6 5 1 -1.5 0.5 0.5 0 0 — S_(2) 0 40 0 -2 12 2 1 0 — S_(3) 0 33 0 (5) -5 -1 0 1 (33)/(5)=6.6 rarr Z=-30 Z_(j) -6 9 -3 -3 0 0 C_(j)-Z_(j) 0 -11 uarr -2 3 0 0 | Iteration-2 | | $C_j$ | -6 | -2 | -5 | 0 | 0 | 0 | | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $B$ | $C_B$ | $X_B$ | $x_1$ | $x_2$ | $x_3$ | $S_1$ | $S_2$ | $S_3$ | $\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_2}\end{array}$ | | $x_1$ | -6 | 5 | 1 | -1.5 | 0.5 | 0.5 | 0 | 0 | — | | $S_2$ | 0 | 40 | 0 | -2 | 12 | 2 | 1 | 0 | — | | $S_3$ | 0 | 33 | 0 | (5) | -5 | -1 | 0 | 1 | $\frac{33}{5}=6.6 \rightarrow$ | | $Z=-\mathbf{3 0}$ | | $Z_j$ | -6 | 9 | -3 | -3 | 0 | 0 | | | | | $C_j-Z_j$ | 0 | $-11 \uparrow$ | -2 | 3 | 0 | 0 | |
Negative minimum C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj is -11 and its column index is 2 . So, the entering variable is x 2 x 2 x_(2)x_2x2.
Minimum ratio is 6.6 and its row index is 3 . So, the leaving basis variable is S 3 S 3 S_(3)S_3S3.
∴ ∴ :.\therefore∴ The pivot element is 5 .
Entering = x 2 = x 2 =x_(2)=x_2=x2, Departing = S 3 = S 3 =S_(3)=S_3=S3, Key Element = 5 = 5 =5=5=5
R 3 ( new ) = R 3 ( old ) ÷ 5 R 1 ( new ) = R 1 ( old ) + 1.5 R 3 (new ) R 2 ( new ) = R 2 ( old ) + 2 R 3 ( new ) R 3 (  new  ) = R 3 (  old  ) ÷ 5 R 1 (  new  ) = R 1 (  old  ) + 1.5 R 3  (new  R 2 (  new  ) = R 2 (  old  ) + 2 R 3 (  new  ) {:[R_(3)(” new “)=R_(3)(” old “)-:5],[{:R_(1)(” new “)=R_(1)(” old “)+1.5R_(3)” (new “)],[R_(2)(” new “)=R_(2)(” old “)+2R_(3)(” new “)]:}\begin{aligned} & R_3(\text { new })=R_3(\text { old }) \div 5 \\ & \left.R_1(\text { new })=R_1(\text { old })+1.5 R_3 \text { (new }\right) \\ & R_2(\text { new })=R_2(\text { old })+2 R_3(\text { new }) \end{aligned}R3( new )=R3( old )÷5R1( new )=R1( old )+1.5R3 (new )R2( new )=R2( old )+2R3( new )
Iteration-3 C j C j C_(j)C_jCj-6-2-5000
B B BBB C B C B C_(B)C_BCB X B X B X_(B)X_BXB x 1 x 1 x_(1)x_1x1 x 2 x 2 x_(2)x_2x2 x 3 x 3 x_(3)x_3x3 s 1 s 1 s_(1)s_1s1 S 2 S 2 S_(2)S_2S2 S 3 S 3 S_(3)S_3S3 MinRatio X B x 3  MinRatio  X B x 3 {:[” MinRatio “],[(X_(B))/(x_(3))]:}\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_3}\end{array} MinRatio XBx3
x 1 x 1 x_(1)x_1x1-614.910-10.200.3—
s 2 s 2 s_(2)s_2s2053.200(10)1.610.4 53.2 10 = 5.32 → 53.2 10 = 5.32 → (53.2)/(10)=5.32 rarr\frac{53.2}{10}=5.32 \rightarrow53.210=5.32→
x 2 x 2 x_(2)x_2x2-26.601-1-0.200.2—
Z = − 102.6 Z = − 102.6 Z=-102.6Z=-102.6Z=−102.6 Z j Z j Z_(j)Z_jZj-6-28-0.80-2.2
C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj00 − 13 ↑ − 13 ↑ -13 uarr-13 \uparrow−13↑0.802.2
Iteration-3 C_(j) -6 -2 -5 0 0 0 B C_(B) X_(B) x_(1) x_(2) x_(3) s_(1) S_(2) S_(3) ” MinRatio (X_(B))/(x_(3))” x_(1) -6 14.9 1 0 -1 0.2 0 0.3 — s_(2) 0 53.2 0 0 (10) 1.6 1 0.4 (53.2)/(10)=5.32 rarr x_(2) -2 6.6 0 1 -1 -0.2 0 0.2 — Z=-102.6 Z_(j) -6 -2 8 -0.8 0 -2.2 C_(j)-Z_(j) 0 0 -13 uarr 0.8 0 2.2 | Iteration-3 | | $C_j$ | -6 | -2 | -5 | 0 | 0 | 0 | | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $B$ | $C_B$ | $X_B$ | $x_1$ | $x_2$ | $x_3$ | $s_1$ | $S_2$ | $S_3$ | $\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_3}\end{array}$ | | $x_1$ | -6 | 14.9 | 1 | 0 | -1 | 0.2 | 0 | 0.3 | — | | $s_2$ | 0 | 53.2 | 0 | 0 | (10) | 1.6 | 1 | 0.4 | $\frac{53.2}{10}=5.32 \rightarrow$ | | $x_2$ | -2 | 6.6 | 0 | 1 | -1 | -0.2 | 0 | 0.2 | — | | $Z=-102.6$ | | $Z_j$ | -6 | -2 | 8 | -0.8 | 0 | -2.2 | | | | | $C_j-Z_j$ | 0 | 0 | $-13 \uparrow$ | 0.8 | 0 | 2.2 | |
Negative minimum C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj is -13 and its column index is 3 . So, the entering variable is x 3 x 3 x_(3)x_3x3.
Minimum ratio is 5.32 and its row index is 2 . So, the leaving basis variable is S 2 S 2 S_(2)S_2S2.
∴ ∴ :.\therefore∴ The pivot element is 10 .
Entering = x 3 = x 3 =x_(3)=x_3=x3, Departing = S 2 = S 2 =S_(2)=S_2=S2, Key Element = 10 = 10 =10=10=10
R 2 ( new ) = R 2 ( old ) ÷ 10 R 1 ( new ) = R 1 ( old ) + R 2 ( new ) R 3 ( new ) = R 3 ( old ) + R 2 ( new ) R 2 (  new  ) = R 2 (  old  ) ÷ 10 R 1 (  new  ) = R 1 (  old  ) + R 2 (  new  ) R 3 (  new  ) = R 3 (  old  ) + R 2 (  new  ) {:[R_(2)(” new “)=R_(2)(” old “)-:10],[R_(1)(” new “)=R_(1)(” old “)+R_(2)(” new “)],[R_(3)(” new “)=R_(3)(” old “)+R_(2)(” new “)]:}\begin{aligned} & R_2(\text { new })=R_2(\text { old }) \div 10 \\ & R_1(\text { new })=R_1(\text { old })+R_2(\text { new }) \\ & R_3(\text { new })=R_3(\text { old })+R_2(\text { new }) \end{aligned}R2( new )=R2( old )÷10R1( new )=R1( old )+R2( new )R3( new )=R3( old )+R2( new )
Iteration-4 C j C j C_(j)C_jCj-6-2-5000
B B B\boldsymbol{B}B C B C B C_(B)\boldsymbol{C}_{\boldsymbol{B}}CB X B X B X_(B)\boldsymbol{X}_{\boldsymbol{B}}XB x 1 x 1 x_(1)\boldsymbol{x}_{\mathbf{1}}x1 x 2 x 2 x_(2)\boldsymbol{x}_{\mathbf{2}}x2 x 3 x 3 x_(3)\boldsymbol{x}_{\mathbf{3}}x3 S 1 S 1 S_(1)\boldsymbol{S}_{\mathbf{1}}S1 S 2 S 2 S_(2)\boldsymbol{S}_{\mathbf{2}}S2 S 3 S 3 S_(3)\boldsymbol{S}_{\mathbf{3}}S3MinRatio
x 1 x 1 x_(1)x_1x1-620.221000.360.10.34
x 3 x 3 x_(3)x_3x3-55.320010.160.10.04
x 2 x 2 x_(2)x_2x2-211.92010-0.040.10.24
Z = − 1 7 1 . 7 6 Z = − 1 7 1 . 7 6 Z=-171.76\boldsymbol{Z}=-\mathbf{1 7 1 . 7 6}Z=−171.76 Z j Z j Z_(j)\boldsymbol{Z}_{\boldsymbol{j}}Zj − 6 − 6 -6\mathbf{- 6}−6 − 2 − 2 -2\mathbf{- 2}−2 − 5 − 5 -5\mathbf{- 5}−5 − 2 . 8 8 − 2 . 8 8 -2.88\mathbf{- 2 . 8 8}−2.88 − 1 . 3 − 1 . 3 -1.3\mathbf{- 1 . 3}−1.3 − 2 . 7 2 − 2 . 7 2 -2.72\mathbf{- 2 . 7 2}−2.72
C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj0002.881.32.72
Iteration-4 C_(j) -6 -2 -5 0 0 0 B C_(B) X_(B) x_(1) x_(2) x_(3) S_(1) S_(2) S_(3) MinRatio x_(1) -6 20.22 1 0 0 0.36 0.1 0.34 x_(3) -5 5.32 0 0 1 0.16 0.1 0.04 x_(2) -2 11.92 0 1 0 -0.04 0.1 0.24 Z=-171.76 Z_(j) -6 -2 -5 -2.88 -1.3 -2.72 C_(j)-Z_(j) 0 0 0 2.88 1.3 2.72 | Iteration-4 | | $C_j$ | -6 | -2 | -5 | 0 | 0 | 0 | | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $\boldsymbol{B}$ | $\boldsymbol{C}_{\boldsymbol{B}}$ | $\boldsymbol{X}_{\boldsymbol{B}}$ | $\boldsymbol{x}_{\mathbf{1}}$ | $\boldsymbol{x}_{\mathbf{2}}$ | $\boldsymbol{x}_{\mathbf{3}}$ | $\boldsymbol{S}_{\mathbf{1}}$ | $\boldsymbol{S}_{\mathbf{2}}$ | $\boldsymbol{S}_{\mathbf{3}}$ | MinRatio | | $x_1$ | -6 | 20.22 | 1 | 0 | 0 | 0.36 | 0.1 | 0.34 | | | $x_3$ | -5 | 5.32 | 0 | 0 | 1 | 0.16 | 0.1 | 0.04 | | | $x_2$ | -2 | 11.92 | 0 | 1 | 0 | -0.04 | 0.1 | 0.24 | | | $\boldsymbol{Z}=-\mathbf{1 7 1 . 7 6}$ | | $\boldsymbol{Z}_{\boldsymbol{j}}$ | $\mathbf{- 6}$ | $\mathbf{- 2}$ | $\mathbf{- 5}$ | $\mathbf{- 2 . 8 8}$ | $\mathbf{- 1 . 3}$ | $\mathbf{- 2 . 7 2}$ | | | | | $C_j-Z_j$ | 0 | 0 | 0 | 2.88 | 1.3 | 2.72 | |
Since all C j − Z j ≥ 0 C j − Z j ≥ 0 C_(j)-Z_(j) >= 0C_j-Z_j \geq 0Cj−Zj≥0
Hence, optimal solution is arrived with value of variables as :
x 1 = 20.22 , x 2 = 11.92 , x 3 = 5.32 x 1 = 20.22 , x 2 = 11.92 , x 3 = 5.32 x_(1)=20.22,x_(2)=11.92,x_(3)=5.32x_1=20.22, x_2=11.92, x_3=5.32x1=20.22,x2=11.92,x3=5.32
Min Z ∗ = − 171.76 Min Z ∗ = − 171.76 MinZ^(**)=-171.76\operatorname{Min} Z^*=-171.76MinZ∗=−171.76
Hence, Min Z = Min Z ∗ − 29 Min Z = Min Z ∗ − 29 Min Z=MinZ^(**)-29\operatorname{Min} Z=\operatorname{Min} Z^*-29MinZ=MinZ∗−29
Min Z = − 171.76 − 29 = − 200.76 Min Z = − 171.76 − 29 = − 200.76 Min Z=-171.76-29=-200.76\operatorname{Min} Z=-171.76-29=-200.76MinZ=−171.76−29=−200.76

Conclusion

The optimal solution to the given LPP is x 1 = 20.22 , x 2 = 11.92 , x 3 = 5.32 x 1 = 20.22 , x 2 = 11.92 , x 3 = 5.32 x_(1)=20.22,x_(2)=11.92,x_(3)=5.32x_1 = 20.22, x_2 = 11.92, x_3 = 5.32x1=20.22,x2=11.92,x3=5.32 with a minimum value of Z = − 200.76 Z = − 200.76 Z=-200.76Z = -200.76Z=−200.76. All C j − Z j C j − Z j C_(j)-Z_(j)C_j – Z_jCj−Zj values are non-negative, confirming that the solution is optimal.
3.(c) यदि u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u=tan^(-1)((x^(3)+y^(3))/(x-y)),x!=yu=\tan ^{-1} \frac{x^3+y^3}{x-y}, x \neq yu=tan−1⁡x3+y3x−y,x≠y
तब दर्शाइए कि x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ 2 u x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = 1 − 4 sin 2 ⁡ u sin ⁡ 2 u x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(1-4sin^(2)u)sin 2ux^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\left(1-4 \sin ^2 u\right) \sin 2 ux2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=(1−4sin2⁡u)sin⁡2u
If u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u = tan − 1 ⁡ x 3 + y 3 x − y , x ≠ y u=tan^(-1)((x^(3)+y^(3))/(x-y)),x!=yu=\tan ^{-1} \frac{x^3+y^3}{x-y}, x \neq yu=tan−1⁡x3+y3x−y,x≠y
then show that x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ 2 u x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = 1 − 4 sin 2 ⁡ u sin ⁡ 2 u x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(1-4sin^(2)u)sin 2ux^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\left(1-4 \sin ^2 u\right) \sin 2 ux2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=(1−4sin2⁡u)sin⁡2u
Answer:
Introduction:
Here, we explore the nature of a function u = tan − 1 ⁡ x 3 + y 3 x − y u = tan − 1 ⁡ x 3 + y 3 x − y u=tan^(-1)((x^(3)+y^(3))/(x-y))u = \tan^{-1}\frac{x^3+y^3}{x-y}u=tan−1⁡x3+y3x−y and investigate its homogeneity.
Homogeneity of u u uuu:
We start by considering whether u u uuu is a homogeneous function.
However, we express tan ⁡ u = x 3 + y 3 x − y tan ⁡ u = x 3 + y 3 x − y tan u=(x^(3)+y^(3))/(x-y)\tan u = \frac{x^3+y^3}{x-y}tan⁡u=x3+y3x−y as z z zzz, where z z zzz is a new variable. Thus, we obtain:
tan ⁡ u = x 3 + y 3 x − y = z → ( 1 ) ⇒ z = x 2 [ 1 + ( y x ) 3 1 − ( y x ) ] tan ⁡ u = x 3 + y 3 x − y = z → ( 1 ) ⇒ z = x 2 1 + y x 3 1 − y x {:[tan u=(x^(3)+y^(3))/(x-y)=z quad rarr(1)],[=>z=x^(2)[(1+((y)/(x))^(3))/(1-((y)/(x)))]]:}\begin{aligned} & \tan u = \frac{x^3+y^3}{x-y} = z \quad \rightarrow(1) \\ & \Rightarrow z = x^2\left[\frac{1+\left(\frac{y}{x}\right)^3}{1-\left(\frac{y}{x}\right)}\right] \end{aligned}tan⁡u=x3+y3x−y=z→(1)⇒z=x2[1+(yx)31−(yx)]
Now, z z zzz is shown to be a homogeneous function of x x xxx and y y yyy of degree 2:
x ∂ z ∂ x + y ∂ z ∂ y = 2 z → ( 2 ) x ∂ z ∂ x + y ∂ z ∂ y = 2 z → ( 2 ) x(del z)/(del x)+y(del z)/(del y)=2z quad rarr(2)x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 2z \quad \rightarrow(2)x∂z∂x+y∂z∂y=2z→(2)
Derivatives of z z zzz and u u uuu:
Next, we calculate the derivatives of z z zzz and u u uuu with respect to x x xxx and y y yyy. These derivatives play a crucial role in our analysis. From equation (1), we have:
∂ z ∂ x = sec 2 ⁡ u ∂ u ∂ x ∂ z ∂ y = sec 2 ⁡ u ∂ u ∂ y → ( 3 ) ∂ z ∂ x = sec 2 ⁡ u ∂ u ∂ x ∂ z ∂ y = sec 2 ⁡ u ∂ u ∂ y → ( 3 ) {:[(del z)/(del x)=sec^(2)u(del u)/(del x)],[(del z)/(del y)=sec^(2)u(del u)/(del y)quad rarr(3)]:}\begin{aligned} & \frac{\partial z}{\partial x} = \sec^2 u \frac{\partial u}{\partial x} \\ & \frac{\partial z}{\partial y} = \sec^2 u \frac{\partial u}{\partial y} \quad \rightarrow(3) \end{aligned}∂z∂x=sec2⁡u∂u∂x∂z∂y=sec2⁡u∂u∂y→(3)
Simplification and Relationships:
Now, we use the derivatives to simplify and establish relationships between x , y , z , u x , y , z , u x,y,z,ux, y, z, ux,y,z,u:
From equation (2), we have:
x ∂ u ∂ x + y ∂ u ∂ y = 2 sin ⁡ u cos ⁡ u ⋅ cos 2 ⁡ u = sin ⁡ 2 u → ( 4 ) x ∂ u ∂ x + y ∂ u ∂ y = 2 sin ⁡ u cos ⁡ u ⋅ cos 2 ⁡ u = sin ⁡ 2 u → ( 4 ) x(del u)/(del x)+y(del u)/(del y)=(2sin u)/(cos u)*cos^(2)u=sin 2u quad rarr(4)x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = \frac{2 \sin u}{\cos u} \cdot \cos^2 u = \sin 2u \quad \rightarrow(4)x∂u∂x+y∂u∂y=2sin⁡ucos⁡u⋅cos2⁡u=sin⁡2u→(4)
From the derivatives in equation (3), we further derive:
∂ 2 z ∂ x 2 = sec 2 ⁡ u ∂ 2 u ∂ x 2 + 2 sec 2 ⁡ u tan ⁡ u ( ∂ u ∂ x ) 2 ∂ 2 z ∂ y 2 = sec 2 ⁡ u ∂ 2 u ∂ y 2 + 2 sec 2 ⁡ u tan ⁡ u ( ∂ u ∂ y ) 2 ∂ 2 z ∂ x 2 = sec 2 ⁡ u ∂ 2 u ∂ x 2 + 2 sec 2 ⁡ u tan ⁡ u ∂ u ∂ x 2 ∂ 2 z ∂ y 2 = sec 2 ⁡ u ∂ 2 u ∂ y 2 + 2 sec 2 ⁡ u tan ⁡ u ∂ u ∂ y 2 {:[(del^(2)z)/(delx^(2))=sec^(2)u(del^(2)u)/(delx^(2))+2sec^(2)u tan u((del u)/(del x))^(2)],[(del^(2)z)/(dely^(2))=sec^(2)u(del^(2)u)/(dely^(2))+2sec^(2)u tan u((del u)/(del y))^(2)]:}\begin{aligned} & \frac{\partial^2 z}{\partial x^2} = \sec^2 u \frac{\partial^2 u}{\partial x^2} + 2 \sec^2 u \tan u \left(\frac{\partial u}{\partial x}\right)^2 \\ & \frac{\partial^2 z}{\partial y^2} = \sec^2 u \frac{\partial^2 u}{\partial y^2} + 2 \sec^2 u \tan u \left(\frac{\partial u}{\partial y}\right)^2 \end{aligned}∂2z∂x2=sec2⁡u∂2u∂x2+2sec2⁡utan⁡u(∂u∂x)2∂2z∂y2=sec2⁡u∂2u∂y2+2sec2⁡utan⁡u(∂u∂y)2
And the mixed partial derivative:
∂ 2 z ∂ x ∂ y = sec 2 ⁡ u ∂ 2 u ∂ x ∂ y + 2 sec 2 ⁡ u tan ⁡ u ∂ u ∂ x ∂ u ∂ y ∂ 2 z ∂ x ∂ y = sec 2 ⁡ u ∂ 2 u ∂ x ∂ y + 2 sec 2 ⁡ u tan ⁡ u ∂ u ∂ x ∂ u ∂ y (del^(2)z)/(del x del y)=sec^(2)u(del^(2)u)/(del x del y)+2sec^(2)u tan u(del u)/(del x)(del u)/(del y)\frac{\partial^2 z}{\partial x \partial y} = \sec^2 u \frac{\partial^2 u}{\partial x \partial y} + 2 \sec^2 u \tan u \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}∂2z∂x∂y=sec2⁡u∂2u∂x∂y+2sec2⁡utan⁡u∂u∂x∂u∂y
Euler’s Theorem and Further Simplification:
By a corollary of Euler’s theorem, we reach the following relationships:
x 2 ∂ 2 z ∂ x 2 + 2 x y ∂ 2 z ∂ x ∂ y + y 2 ∂ 2 z ∂ y 2 = 2 ( 2 − 1 ) z ⇒ sec 2 ⁡ u ( x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 ) + 2 sec 2 ⁡ u tan ⁡ u [ x 2 ( ∂ u ∂ x ) 2 + 2 x y ∂ u ∂ x ∂ u ∂ y + y 2 ( ∂ u ∂ y ) 2 ] = 2 tan ⁡ u x 2 ∂ 2 z ∂ x 2 + 2 x y ∂ 2 z ∂ x ∂ y + y 2 ∂ 2 z ∂ y 2 = 2 ( 2 − 1 ) z ⇒ sec 2 ⁡ u x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 + 2 sec 2 ⁡ u tan ⁡ u x 2 ∂ u ∂ x 2 + 2 x y ∂ u ∂ x ∂ u ∂ y + y 2 ∂ u ∂ y 2 = 2 tan ⁡ u {:[x^(2)(del^(2)z)/(delx^(2))+2xy(del^(2)z)/(del x del y)+y^(2)(del^(2)z)/(dely^(2))=2(2-1)z],[=>sec^(2)u(x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2)))+2sec^(2)u tan u[x^(2)((del u)/(del x))^(2)+2xy(del u)/(del x)(del u)/(del y)+y^(2)((del u)/(del y))^(2)]=2tan u]:}\begin{aligned} & x^2 \frac{\partial^2 z}{\partial x^2} + 2xy \frac{\partial^2 z}{\partial x \partial y} + y^2 \frac{\partial^2 z}{\partial y^2} = 2(2-1)z \\ & \Rightarrow \sec^2 u \left(x^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2}\right) + 2 \sec^2 u \tan u \left[x^2\left(\frac{\partial u}{\partial x}\right)^2+2xy \frac{\partial u}{\partial x} \frac{\partial u}{\partial y}+y^2\left(\frac{\partial u}{\partial y}\right)^2\right] = 2 \tan u \end{aligned}x2∂2z∂x2+2xy∂2z∂x∂y+y2∂2z∂y2=2(2−1)z⇒sec2⁡u(x2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2)+2sec2⁡utan⁡u[x2(∂u∂x)2+2xy∂u∂x∂u∂y+y2(∂u∂y)2]=2tan⁡u
divide by s e c 2 u s e c 2 u sec^(2)usec^2 usec2u
⇒ x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 + 2 tan ⁡ u ( x ∂ u ∂ x + y ∂ u ∂ y ) 2 = 2 sin ⁡ u cos ⁡ u ⇒ x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = sin ⁡ 2 u − 2 tan ⁡ u sin 2 ⁡ 2 u ( b y ( 4 ) ) = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ 2 u ⇒ x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 + 2 tan ⁡ u x ∂ u ∂ x + y ∂ u ∂ y 2 = 2 sin ⁡ u cos ⁡ u ⇒ x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = sin ⁡ 2 u − 2 tan ⁡ u sin 2 ⁡ 2 u ( b y ( 4 ) ) = 1 − 4 sin 2 ⁡ u sin ⁡ 2 u {:[=>x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))+2tan u(x(del u)/(del x)+y(del u)/(del y))^(2)=2sin u cos u],[=>x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=sin 2u-2tan usin^(2)2u(by(4))],[=(1-4sin^(2)u)sin 2u]:}\begin{aligned} & \Rightarrow x^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}+2 \tan u\left(x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\right)^2=2 \sin u \cos u \\ & \Rightarrow x^2 \frac{\partial^2 u}{\partial x^2}+2 x y \frac{\partial^2 u}{\partial x \partial y}+y^2 \frac{\partial^2 u}{\partial y^2}=\sin 2u-2 \tan u \sin ^2 2u(b y(4)) \\ & =\left(1-4 \sin ^2 u\right) \sin 2 u \end{aligned}⇒x2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2+2tan⁡u(x∂u∂x+y∂u∂y)2=2sin⁡ucos⁡u⇒x2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=sin⁡2u−2tan⁡usin2⁡2u(by(4))=(1−4sin2⁡u)sin⁡2u
Final Relationship:
Finally, we obtain:
x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = ( 1 − 4 sin 2 ⁡ u ) sin ⁡ 2 u x 2 ∂ 2 u ∂ x 2 + 2 x y ∂ 2 u ∂ x ∂ y + y 2 ∂ 2 u ∂ y 2 = 1 − 4 sin 2 ⁡ u sin ⁡ 2 u x^(2)(del^(2)u)/(delx^(2))+2xy(del^(2)u)/(del x del y)+y^(2)(del^(2)u)/(dely^(2))=(1-4sin^(2)u)sin 2ux^2 \frac{\partial^2 u}{\partial x^2} + 2xy \frac{\partial^2 u}{\partial x \partial y} + y^2 \frac{\partial^2 u}{\partial y^2} =\left(1-4 \sin ^2 u\right) \sin 2 ux2∂2u∂x2+2xy∂2u∂x∂y+y2∂2u∂y2=(1−4sin2⁡u)sin⁡2u
This equation represents a significant relationship in our analysis.
4.(a) यदि v ( r , θ ) = ( r − 1 r ) sin ⁡ θ , r ≠ 0 v ( r , θ ) = r − 1 r sin ⁡ θ , r ≠ 0 v(r,theta)=(r-(1)/(r))sin theta,r!=0v(r, \theta)=\left(r-\frac{1}{r}\right) \sin \theta, r \neq 0v(r,θ)=(r−1r)sin⁡θ,r≠0,
तब विश्लेषिक फलन f ( z ) = u ( r , θ ) + i v ( r , θ ) f ( z ) = u ( r , θ ) + i v ( r , θ ) f(z)=u(r,theta)+iv(r,theta)f(z)=u(r, \theta)+i v(r, \theta)f(z)=u(r,θ)+iv(r,θ) ज्ञात कीजिए ।
If v ( r , θ ) = ( r − 1 r ) sin ⁡ θ , r ≠ 0 v ( r , θ ) = r − 1 r sin ⁡ θ , r ≠ 0 v(r,theta)=(r-(1)/(r))sin theta,r!=0v(r, \theta)=\left(r-\frac{1}{r}\right) \sin \theta, r \neq 0v(r,θ)=(r−1r)sin⁡θ,r≠0,
then find an analytic function f ( z ) = u ( r , θ ) + i v ( r , θ ) f ( z ) = u ( r , θ ) + i v ( r , θ ) f(z)=u(r,theta)+iv(r,theta)f(z)=u(r, \theta)+i v(r, \theta)f(z)=u(r,θ)+iv(r,θ)
Answer:
Introduction:
The problem involves finding an analytic function f ( z ) f ( z ) f(z)f(z)f(z) given a complex-valued function v ( r , θ ) v ( r , θ ) v(r,theta)v(r, \theta)v(r,θ) in polar coordinates.
Solution:
Step 1: Cauchy-Riemann Equations in Polar Coordinates
The Cauchy-Riemann equations in polar coordinates are as follows:
∂ u ∂ r = 1 r ∂ v ∂ θ (1) 1 r ∂ u ∂ θ = − ∂ v ∂ r (2) ∂ u ∂ r = 1 r ∂ v ∂ θ (1) 1 r ∂ u ∂ θ = − ∂ v ∂ r (2) {:[(del u)/(del r)=(1)/(r)(del v)/(del theta)quad(1)],[(1)/(r)(del u)/(del theta)=-(del v)/(del r)quad(2)]:}\begin{aligned} & \frac{\partial u}{\partial r}=\frac{1}{r} \frac{\partial v}{\partial \theta} \quad \text{(1)} \\ & \frac{1}{r} \frac{\partial u}{\partial \theta}=-\frac{\partial v}{\partial r} \quad \text{(2)} \end{aligned}∂u∂r=1r∂v∂θ(1)1r∂u∂θ=−∂v∂r(2)
Step 2: Expressions for Partial Derivatives of v v vvv
From the given function v ( r , θ ) v ( r , θ ) v(r,theta)v(r, \theta)v(r,θ), we have the following expressions for the partial derivatives:
∂ v ∂ r = ( 1 + 1 r 2 ) sin ⁡ θ (3a) ∂ v ∂ θ = ( r − 1 r ) cos ⁡ θ (3b) ∂ v ∂ r = 1 + 1 r 2 sin ⁡ θ (3a) ∂ v ∂ θ = r − 1 r cos ⁡ θ (3b) {:[(del v)/(del r)=(1+(1)/(r^(2)))sin thetaquad(3a)],[(del v)/(del theta)=(r-(1)/(r))cos thetaquad(3b)]:}\begin{aligned} & \frac{\partial v}{\partial r}=\left(1+\frac{1}{r^2}\right) \sin \theta \quad \text{(3a)} \\ & \frac{\partial v}{\partial \theta}=\left(r-\frac{1}{r}\right) \cos \theta \quad \text{(3b)} \end{aligned}∂v∂r=(1+1r2)sin⁡θ(3a)∂v∂θ=(r−1r)cos⁡θ(3b)
Step 3: Solve for Partial Derivatives of u u uuu
Using equations (1) and (3), we can calculate the partial derivatives of u u uuu as follows:
∂ u ∂ r = 1 r [ r − 1 r ] cos ⁡ θ = ( 1 − 1 r 2 ) cos ⁡ θ (4) ∂ u ∂ θ = − r ∂ v ∂ r = − r [ 1 + 1 r 2 ] sin ⁡ θ = [ − r − 1 r ] sin ⁡ θ (5) ∂ u ∂ r = 1 r r − 1 r cos ⁡ θ = 1 − 1 r 2 cos ⁡ θ (4) ∂ u ∂ θ = − r ∂ v ∂ r = − r 1 + 1 r 2 sin ⁡ θ = − r − 1 r sin ⁡ θ (5) {:[(del u)/(del r)=(1)/(r)[r-(1)/(r)]cos theta=(1-(1)/(r^(2)))cos thetaquad(4)],[(del u)/(del theta)=-r(del v)/(del r)=-r[1+(1)/(r^(2))]sin theta=[-r-(1)/(r)]sin thetaquad(5)]:}\begin{aligned} & \frac{\partial u}{\partial r}=\frac{1}{r}\left[r-\frac{1}{r}\right] \cos \theta=\left(1-\frac{1}{r^2}\right) \cos \theta \quad \text{(4)} \\ & \frac{\partial u}{\partial \theta}=-r \frac{\partial v}{\partial r}=-r\left[1+\frac{1}{r^2}\right] \sin \theta=\left[-r-\frac{1}{r}\right] \sin \theta \quad \text{(5)} \end{aligned}∂u∂r=1r[r−1r]cos⁡θ=(1−1r2)cos⁡θ(4)∂u∂θ=−r∂v∂r=−r[1+1r2]sin⁡θ=[−r−1r]sin⁡θ(5)
Step 4: Integrate Partial Derivatives
Integrating equation (4) with respect to r r rrr and equation (5) with respect to θ θ theta\thetaθ, we obtain:
u ( r , θ ) = ( r + 1 r ) cos ⁡ θ + f ( θ ) (6) u ( r , θ ) = ( r + 1 r ) cos ⁡ θ + f ( r ) (7) u ( r , θ ) = r + 1 r cos ⁡ θ + f ( θ ) (6) u ( r , θ ) = r + 1 r cos ⁡ θ + f ( r ) (7) {:[u(r”,”theta)=(r+(1)/(r))cos theta+f(theta)quad(6)],[u(r”,”theta)=(r+(1)/(r))cos theta+f(r)quad(7)]:}\begin{aligned} & u(r, \theta)=\left(r+\frac{1}{r}\right) \cos \theta+f(\theta) \quad \text{(6)} \\ & u(r, \theta)=\left(r+\frac{1}{r}\right) \cos \theta+f{(r)} \quad \text{(7)} \end{aligned}u(r,θ)=(r+1r)cos⁡θ+f(θ)(6)u(r,θ)=(r+1r)cos⁡θ+f(r)(7)
Step 5: Determine f ( θ ) f ( θ ) f(theta)f(\theta)f(θ) and f ( r ) f ( r ) f(r)f{(r)}f(r)
Equating equations (6) and (7), we find:
f ( θ ) = 0 = f ( r ) u ( r , θ ) = ( r + 1 r ) cos ⁡ θ f ( θ ) = 0 = f ( r ) u ( r , θ ) = r + 1 r cos ⁡ θ {:[f(theta)=0=f(r)],[u(r”,”theta)=(r+(1)/(r))cos theta]:}\begin{aligned} & f(\theta)=0=f{(r)} \\ & u(r, \theta)=\left(r+\frac{1}{r}\right) \cos \theta \end{aligned}f(θ)=0=f(r)u(r,θ)=(r+1r)cos⁡θ
Conclusion:
Therefore, the analytic function f ( z ) f ( z ) f(z)f(z)f(z) is given by:
f ( z ) = ( r + 1 r ) cos ⁡ θ + i [ r − 1 r ] sin ⁡ θ f ( z ) = r + 1 r cos ⁡ θ + i r − 1 r sin ⁡ θ f(z)=(r+(1)/(r))cos theta+i[r-(1)/(r)]sin thetaf(z)=\left(r+\frac{1}{r}\right) \cos \theta+i\left[r-\frac{1}{r}\right] \sin \thetaf(z)=(r+1r)cos⁡θ+i[r−1r]sin⁡θ
4.(b) दर्शाइए कि ∫ 0 π / 2 sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) ∫ 0 π / 2   sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) int_(0)^(pi//2)(sin^(2)x)/(sin x+cos x)dx=(1)/(sqrt2)log _(e)(1+sqrt2)\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x=\frac{1}{\sqrt{2}} \log _e(1+\sqrt{2})∫0π/2sin2⁡xsin⁡x+cos⁡xdx=12loge⁡(1+2)
Show that ∫ 0 π / 2 sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) ∫ 0 π / 2   sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x = 1 2 log e ⁡ ( 1 + 2 ) int_(0)^(pi//2)(sin^(2)x)/(sin x+cos x)dx=(1)/(sqrt2)log _(e)(1+sqrt2)\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x=\frac{1}{\sqrt{2}} \log _e(1+\sqrt{2})∫0π/2sin2⁡xsin⁡x+cos⁡xdx=12loge⁡(1+2)
Answer:
Step 1: Define the Integral
We start by defining the integral I I III as:
I = ∫ 0 π / 2 sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x I = ∫ 0 π / 2   sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x I=int_(0)^(pi//2)(sin^(2)x)/(sin x+cos x)dxI=\int_0^{\pi / 2} \frac{\sin^2 x}{\sin x+\cos x} dxI=∫0π/2sin2⁡xsin⁡x+cos⁡xdx
Step 2: Apply Property of Definite Integrals
By using the property ∫ 0 a f ( x ) = ∫ 0 a f ( a − x ) ∫ 0 a   f ( x ) = ∫ 0 a   f ( a − x ) int_(0)^(a)f(x)=int_(0)^(a)f(a-x)\int_0^a f(x) = \int_0^a f(a-x)∫0af(x)=∫0af(a−x), we can rewrite the integral as follows:
I = ∫ 0 π / 2 sin 2 ⁡ ( π 2 − x ) sin ⁡ ( π 2 − x ) + cos ⁡ ( π 2 − x ) d x I = ∫ 0 π / 2   sin 2 ⁡ π 2 − x sin ⁡ π 2 − x + cos ⁡ π 2 − x d x I=int_(0)^(pi//2)(sin^(2)((pi)/(2)-x))/(sin((pi)/(2)-x)+cos((pi)/(2)-x))dxI=\int_0^{\pi / 2} \frac{\sin^2\left(\frac{\pi}{2}-x\right)}{\sin\left(\frac{\pi}{2}-x\right)+\cos\left(\frac{\pi}{2}-x\right)} dxI=∫0π/2sin2⁡(π2−x)sin⁡(π2−x)+cos⁡(π2−x)dx
Step 3: Simplify the Integral
Simplifying the integrand further, we get:
I = ∫ 0 π 2 cos 2 ⁡ x cos ⁡ x + sin ⁡ x d x I = ∫ 0 π 2   cos 2 ⁡ x cos ⁡ x + sin ⁡ x d x I=int_(0)^((pi)/(2))(cos^(2)x)/(cos x+sin x)dxI=\int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos x+\sin x} dxI=∫0π2cos2⁡xcos⁡x+sin⁡xdx
Step 4: Express the Integral in a Different Form
Next, we write the integral as:
2 I = ∫ 0 π / 2 1 sin ⁡ x + cos ⁡ x d x = 1 2 ∫ 0 π / 2 1 cos ⁡ ( x − π 4 ) d x 2 I = ∫ 0 π / 2   1 sin ⁡ x + cos ⁡ x d x = 1 2 ∫ 0 π / 2   1 cos ⁡ x − π 4 d x 2I=int_(0)^(pi//2)(1)/(sin x+cos x)dx=(1)/(sqrt2)int_(0)^(pi//2)(1)/(cos(x-(pi)/(4)))dx2I=\int_0^{\pi / 2} \frac{1}{\sin x+\cos x} dx = \frac{1}{\sqrt{2}} \int_0^{\pi / 2} \frac{1}{\cos \left(x-\frac{\pi}{4}\right)} dx2I=∫0π/21sin⁡x+cos⁡xdx=12∫0π/21cos⁡(x−π4)dx
Step 5: Evaluate the Integral
Now, we evaluate the integral:
2 I = 1 2 [ ln ⁡ | sec ⁡ ( x − π 4 ) + tan ⁡ ( x − π 4 ) | ] 0 π / 2 2 I = 1 2 ln ⁡ sec ⁡ x − π 4 + tan ⁡ x − π 4 0 π / 2 2I=(1)/(sqrt2)[ln|sec(x-(pi)/(4))+tan(x-(pi)/(4))|]_(0)^(pi//2)2I = \frac{1}{\sqrt{2}} \left[\ln \left| \sec \left(x-\frac{\pi}{4}\right) + \tan \left(x-\frac{\pi}{4}\right) \right| \right]_0^{\pi / 2}2I=12[ln⁡|sec⁡(x−π4)+tan⁡(x−π4)|]0π/2
Step 6: Calculate the Values
Calculate the values of the integral:
2 I = 1 2 [ ln ⁡ | 2 + 1 | − ln ⁡ | 2 − 1 | ] 2 I = 1 2 ln ⁡ | 2 + 1 | − ln ⁡ | 2 − 1 | 2I=(1)/(sqrt2)[ln |sqrt2+1|-ln |sqrt2-1|]2I = \frac{1}{\sqrt{2}} \left[\ln |\sqrt{2}+1| – \ln |\sqrt{2}-1| \right]2I=12[ln⁡|2+1|−ln⁡|2−1|]
Step 7: Simplify
Simplify the expression:
2 I = 1 2 ln ⁡ | 2 + 1 2 − 1 | 2 I = 1 2 ln ⁡ 2 + 1 2 − 1 2I=(1)/(sqrt2)ln|(sqrt2+1)/(sqrt2-1)|2I = \frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2}+1}{\sqrt{2}-1}\right|2I=12ln⁡|2+12−1|
Step 8: Solve for I I III
Solve for I I III:
I = 2 2 2 ln ⁡ | 2 + 1 | I = 2 2 2 ln ⁡ | 2 + 1 | I=(2)/(2sqrt2)ln |sqrt2+1|I = \frac{2}{2 \sqrt{2}} \ln |\sqrt{2}+1|I=222ln⁡|2+1|
Step 9: Final Result
Finally, we arrive at the result:
I = 1 2 ln ⁡ | 2 + 1 | I = 1 2 ln ⁡ | 2 + 1 | I=(1)/(sqrt2)ln |sqrt2+1|I = \frac{1}{\sqrt{2}} \ln |\sqrt{2}+1|I=12ln⁡|2+1|
Conclusion:
The integral ∫ 0 π / 2 sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x ∫ 0 π / 2   sin 2 ⁡ x sin ⁡ x + cos ⁡ x d x int_(0)^(pi//2)(sin^(2)x)/(sin x+cos x)dx\int_0^{\pi / 2} \frac{\sin^2 x}{\sin x + \cos x} dx∫0π/2sin2⁡xsin⁡x+cos⁡xdx is equal to 1 2 ln ⁡ ( 1 + 2 ) 1 2 ln ⁡ ( 1 + 2 ) (1)/(sqrt2)ln(1+sqrt2)\frac{1}{\sqrt{2}} \ln(1+\sqrt{2})12ln⁡(1+2), as shown.
  1. (c) वोगेल की सम्निकटन विधि से निम्नलिखित परिवहन समस्या का आरंभिक आधारिक सुसंगत हल ज्ञात कीजिए । इस हल का उपयोग कर समस्या का इष्टतम हल एवं परिवहन लागत ज्ञात कीजिए ।
D 1 D 2 D 3 D 4 Supply S 1 10 0 20 11 15 S 2 12 8 9 20 25 S 3 0 14 16 18 10 Demand 5 20 15 10 D 1 D 2 D 3 D 4  Supply  S 1 10 0 20 11 15 S 2 12 8 9 20 25 S 3 0 14 16 18 10  Demand  5 20 15 10 {:[,D_(1),D_(2),D_(3),D_(4),” Supply “],[S_(1),10,0,20,11,15],[S_(2),12,8,9,20,25],[S_(3),0,14,16,18,10],[” Demand “,5,20,15,10,]:}\begin{array}{|l|l|l|l|l||l|} \hline & D_1 & D_2 & D_3 & D_4 & \text { Supply } \\ \hline S_1 & 10 & 0 & 20 & 11 & 15 \\ \hline S_2 & 12 & 8 & 9 & 20 & 25 \\ \hline S_3 & 0 & 14 & 16 & 18 & 10 \\ \hline \hline \text { Demand } & 5 & 20 & 15 & 10 & \\ \hline \end{array}D1D2D3D4 Supply S1100201115S212892025S3014161810 Demand 5201510
Find the initial basic feasible solution of the following transportation problem by Vogel’s approximation method and use it to find the optimal solution and the transportation cost of the problem.
D 1 D 2 D 3 D 4 Supply S 1 10 0 20 11 15 S 2 12 8 9 20 25 S 3 0 14 16 18 10 Demand 5 20 15 10 D 1 D 2 D 3 D 4  Supply  S 1 10 0 20 11 15 S 2 12 8 9 20 25 S 3 0 14 16 18 10  Demand  5 20 15 10 {:[,D_(1),D_(2),D_(3),D_(4),” Supply “],[S_(1),10,0,20,11,15],[S_(2),12,8,9,20,25],[S_(3),0,14,16,18,10],[” Demand “,5,20,15,10,]:}\begin{array}{|l|l|l|l|l||l|} \hline & D_1 & D_2 & D_3 & D_4 & \text { Supply } \\ \hline S_1 & 10 & 0 & 20 & 11 & 15 \\ \hline S_2 & 12 & 8 & 9 & 20 & 25 \\ \hline S_3 & 0 & 14 & 16 & 18 & 10 \\ \hline \hline \text { Demand } & 5 & 20 & 15 & 10 & \\ \hline \end{array}D1D2D3D4 Supply S1100201115S212892025S3014161810 Demand 5201510
Answer:
Initial feasible solution is
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4SupplyRow Penalty
S 1 S 1 S_(1)S_1S110 0 ( 1 5 ) 0 ( 1 5 ) 0(15)0(\mathbf{1 5})0(15)201115 10 | 11 | − | − | − | − | 10 | 11 | − | − | − | − | 10|11|-|-|-|-|10|11|-|-|-|-|10|11|−|−|−|−|
S 2 S 2 S_(2)S_2S212 8 ( 5 ) 8 ( 5 ) 8(5)8(5)8(5) 9 ( 15 ) 9 ( 15 ) 9(15)9(15)9(15) 20 ( 5 ) 20 ( 5 ) 20(5)20(5)20(5)25 1 | 1 | 1 | 12 | 20 | − − | 1 | 1 | 1 | 12 | 20 | − − | 1|1|1|12|20|–|1|1| 1|12| 20|–|1|1|1|12|20|−−|
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5)1416 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10 14 | 2 | 2 | 4 | 18 | 18 | 14 | 2 | 2 | 4 | 18 | 18 | 14|2|2|4|18|18|14|2| 2|4| 18|18|14|2|2|4|18|18|
Demand5201510
10877
Column—877
Penalty—672
———2
——–18
D_(1) D_(2) D_(3) D_(4) Supply Row Penalty S_(1) 10 0(15) 20 11 15 10|11|-|-|-|-| S_(2) 12 8(5) 9(15) 20(5) 25 1|1|1|12|20|–| S_(3) 0(5) 14 16 18(5) 10 14|2|2|4|18|18| Demand 5 20 15 10 10 8 7 7 Column — 8 7 7 Penalty — 6 7 2 — — — 2 — — – 18 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | Row Penalty | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $S_1$ | 10 | $0(\mathbf{1 5})$ | 20 | 11 | 15 | $10\|11\|-\|-\|-\|-\|$ | | $S_2$ | 12 | $8(5)$ | $9(15)$ | $20(5)$ | 25 | $1\|1\| 1\|12\| 20\|–\|$ | | $S_3$ | $0(5)$ | 14 | 16 | $18(5)$ | 10 | $14\|2\| 2\|4\| 18\|18\|$ | | Demand | 5 | 20 | 15 | 10 | | | | | 10 | 8 | 7 | 7 | | | | Column | — | 8 | 7 | 7 | | | | Penalty | — | 6 | 7 | 2 | | | | | — | — | — | 2 | | | | | — | — | – | 18 | | |
The minimum total transportation cost = 0 × 15 + 8 × 5 + 9 × 15 + 20 × 5 + 0 × 5 + 18 × 5 = 365 = 0 × 15 + 8 × 5 + 9 × 15 + 20 × 5 + 0 × 5 + 18 × 5 = 365 =0xx15+8xx5+9xx15+20 xx5+0xx5+18 xx5=365=0 \times 15+8 \times 5+9 \times 15+20 \times 5+0 \times 5+18 \times 5=365=0×15+8×5+9×15+20×5+0×5+18×5=365
Here, the number of allocated cells = 6 = 6 =6=6=6 is equal to m + n − 1 = 3 + 4 − 1 = 6 m + n − 1 = 3 + 4 − 1 = 6 m+n-1=3+4-1=6m+n-1=3+4-1=6m+n−1=3+4−1=6
∴ ∴ :.\therefore∴ This solution is non-degenerate
Optimality test using modi method…
Allocation Table is
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4Supply
S 1 S 1 S_(1)S_1S110 0 ( 15 ) 0 ( 15 ) 0(15)0(15)0(15)201115
S 2 S 2 S_(2)S_2S212 8 ( 5 ) 8 ( 5 ) 8(5)8(5)8(5) 9 ( 15 ) 9 ( 15 ) 9(15)9(15)9(15) 20 ( 5 ) 20 ( 5 ) 20(5)20(5)20(5)25
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5)1416 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10
Demand5201510
D_(1) D_(2) D_(3) D_(4) Supply S_(1) 10 0(15) 20 11 15 S_(2) 12 8(5) 9(15) 20(5) 25 S_(3) 0(5) 14 16 18(5) 10 Demand 5 20 15 10 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | | :— | :— | :— | :— | :— | :— | | $S_1$ | 10 | $0(15)$ | 20 | 11 | 15 | | $S_2$ | 12 | $8(5)$ | $9(15)$ | $20(5)$ | 25 | | $S_3$ | $0(5)$ | 14 | 16 | $18(5)$ | 10 | | Demand | 5 | 20 | 15 | 10 | |
Iteration-1 of optimality test
  1. Find u i u i u_(i)u_iui and v j v j v_(j)v_jvj for all occupied cells ( i , j ) ( i , j ) (i,j)(\mathrm{i}, \mathrm{j})(i,j), where c i j = u i + v j c i j = u i + v j c_(ij)=u_(i)+v_(j)c_{i j}=u_i+v_jcij=ui+vj
  2. Substituting, u 2 = 0 u 2 = 0 u_(2)=0u_2=0u2=0, we get
  3. c 22 = u 2 + v 2 ⇒ v 2 = c 22 − u 2 ⇒ v 2 = 8 − 0 ⇒ v 2 = 8 c 22 = u 2 + v 2 ⇒ v 2 = c 22 − u 2 ⇒ v 2 = 8 − 0 ⇒ v 2 = 8 c_(22)=u_(2)+v_(2)=>v_(2)=c_(22)-u_(2)=>v_(2)=8-0=>v_(2)=8c_{22}=u_2+v_2 \Rightarrow v_2=c_{22}-u_2 \Rightarrow v_2=8-0 \Rightarrow v_2=8c22=u2+v2⇒v2=c22−u2⇒v2=8−0⇒v2=8
  4. c 12 = u 1 + v 2 ⇒ u 1 = c 12 − v 2 ⇒ u 1 = 0 − 8 ⇒ u 1 = − 8 c 12 = u 1 + v 2 ⇒ u 1 = c 12 − v 2 ⇒ u 1 = 0 − 8 ⇒ u 1 = − 8 c_(12)=u_(1)+v_(2)=>u_(1)=c_(12)-v_(2)=>u_(1)=0-8=>u_(1)=-8c_{12}=u_1+v_2 \Rightarrow u_1=c_{12}-v_2 \Rightarrow u_1=0-8 \Rightarrow u_1=-8c12=u1+v2⇒u1=c12−v2⇒u1=0−8⇒u1=−8
  5. c 23 = u 2 + v 3 ⇒ v 3 = c 23 − u 2 ⇒ v 3 = 9 − 0 ⇒ v 3 = 9 c 23 = u 2 + v 3 ⇒ v 3 = c 23 − u 2 ⇒ v 3 = 9 − 0 ⇒ v 3 = 9 c_(23)=u_(2)+v_(3)=>v_(3)=c_(23)-u_(2)=>v_(3)=9-0=>v_(3)=9c_{23}=u_2+v_3 \Rightarrow v_3=c_{23}-u_2 \Rightarrow v_3=9-0 \Rightarrow v_3=9c23=u2+v3⇒v3=c23−u2⇒v3=9−0⇒v3=9
  6. c 24 = u 2 + v 4 ⇒ v 4 = c 24 − u 2 ⇒ v 4 = 20 − 0 ⇒ v 4 = 20 c 24 = u 2 + v 4 ⇒ v 4 = c 24 − u 2 ⇒ v 4 = 20 − 0 ⇒ v 4 = 20 c_(24)=u_(2)+v_(4)=>v_(4)=c_(24)-u_(2)=>v_(4)=20-0=>v_(4)=20c_{24}=u_2+v_4 \Rightarrow v_4=c_{24}-u_2 \Rightarrow v_4=20-0 \Rightarrow v_4=20c24=u2+v4⇒v4=c24−u2⇒v4=20−0⇒v4=20
  7. c 34 = u 3 + v 4 ⇒ u 3 = c 34 − v 4 ⇒ u 3 = 18 − 20 ⇒ u 3 = − 2 c 34 = u 3 + v 4 ⇒ u 3 = c 34 − v 4 ⇒ u 3 = 18 − 20 ⇒ u 3 = − 2 c_(34)=u_(3)+v_(4)=>u_(3)=c_(34)-v_(4)=>u_(3)=18-20=>u_(3)=-2c_{34}=u_3+v_4 \Rightarrow u_3=c_{34}-v_4 \Rightarrow u_3=18-20 \Rightarrow u_3=-2c34=u3+v4⇒u3=c34−v4⇒u3=18−20⇒u3=−2
  8. c 31 = u 3 + v 1 ⇒ v 1 = c 31 − u 3 ⇒ v 1 = 0 + 2 ⇒ v 1 = 2 c 31 = u 3 + v 1 ⇒ v 1 = c 31 − u 3 ⇒ v 1 = 0 + 2 ⇒ v 1 = 2 c_(31)=u_(3)+v_(1)=>v_(1)=c_(31)-u_(3)=>v_(1)=0+2=>v_(1)=2c_{31}=u_3+v_1 \Rightarrow v_1=c_{31}-u_3 \Rightarrow v_1=0+2 \Rightarrow v_1=2c31=u3+v1⇒v1=c31−u3⇒v1=0+2⇒v1=2
D 1 D 2 D 3 D 4 Supply u i S 1 10 0 ( 1 5 ) 20 11 15 u 1 = − 8 S 2 12 8 ( 5 ) 9 ( 1 5 ) 20 ( 5 ) 25 u 2 = 0 S 3 0 ( 5 ) 14 16 18 ( 5 ) 10 u 3 = − 2 Demand 5 20 15 10 v j v 1 = 2 v 2 = 8 v 3 = 9 v 4 = 20 D 1 D 2 D 3 D 4  Supply  u i S 1 10 0 ( 1 5 ) 20 11 15 u 1 = − 8 S 2 12 8 ( 5 ) 9 ( 1 5 ) 20 ( 5 ) 25 u 2 = 0 S 3 0 ( 5 ) 14 16 18 ( 5 ) 10 u 3 = − 2  Demand  5 20 15 10 v j v 1 = 2 v 2 = 8 v 3 = 9 v 4 = 20 {:[,D_(1),D_(2),D_(3),D_(4),” Supply “,u_(i)],[S_(1),10,0(15),20,11,15,u_(1)=-8],[S_(2),12,8(5),9(15),20(5),25,u_(2)=0],[S_(3),0(5),14,16,18(5),10,u_(3)=-2],[” Demand “,5,20,15,10,,],[v_(j),v_(1)=2,v_(2)=8,v_(3)=9,v_(4)=20,,]:}\begin{array}{|l|l|l|l|l||l|l|} \hline & D_1 & D_2 & D_3 & D_4 & \text { Supply } & u_i \\ \hline S_1 & 10 & 0(\mathbf{1 5 )} & 20 & 11 & 15 & u_1=-8 \\ \hline S_2 & 12 & 8(5) & 9(\mathbf{1 5 )} & 20(5) & 25 & u_2=0 \\ \hline S_3 & 0(5) & 14 & 16 & 18(5) & 10 & u_3=-2 \\ \hline \hline \text { Demand } & 5 & 20 & 15 & 10 & & \\ \hline v_j & v_1=2 & v_2=8 & v_3=9 & v_4=20 & & \\ \hline \end{array}D1D2D3D4 Supply uiS1100(15)201115u1=−8S2128(5)9(15)20(5)25u2=0S30(5)141618(5)10u3=−2 Demand 5201510vjv1=2v2=8v3=9v4=20
  1. Find d i j d i j d_(ij)d_{i j}dij for all unoccupied cells(i,j), where d i j = c i j − ( u i + v j ) d i j = c i j − u i + v j d_(ij)=c_(ij)-(u_(i)+v_(j))d_{i j}=c_{i j}-\left(u_i+v_j\right)dij=cij−(ui+vj)
  2. d 11 = c 11 − ( u 1 + v 1 ) = 10 − ( − 8 + 2 ) = 16 d 11 = c 11 − u 1 + v 1 = 10 − ( − 8 + 2 ) = 16 d_(11)=c_(11)-(u_(1)+v_(1))=10-(-8+2)=16d_{11}=c_{11}-\left(u_1+v_1\right)=10-(-8+2)=16d11=c11−(u1+v1)=10−(−8+2)=16
  3. d 13 = c 13 − ( u 1 + v 3 ) = 20 − ( − 8 + 9 ) = 19 d 13 = c 13 − u 1 + v 3 = 20 − ( − 8 + 9 ) = 19 d_(13)=c_(13)-(u_(1)+v_(3))=20-(-8+9)=19d_{13}=c_{13}-\left(u_1+v_3\right)=20-(-8+9)=19d13=c13−(u1+v3)=20−(−8+9)=19
  4. d 14 = c 14 − ( u 1 + v 4 ) = 11 − ( − 8 + 20 ) = − 1 d 14 = c 14 − u 1 + v 4 = 11 − ( − 8 + 20 ) = − 1 d_(14)=c_(14)-(u_(1)+v_(4))=11-(-8+20)=-1d_{14}=c_{14}-\left(u_1+v_4\right)=11-(-8+20)=-1d14=c14−(u1+v4)=11−(−8+20)=−1
  5. d 21 = c 21 − ( u 2 + v 1 ) = 12 − ( 0 + 2 ) = 10 d 21 = c 21 − u 2 + v 1 = 12 − ( 0 + 2 ) = 10 d_(21)=c_(21)-(u_(2)+v_(1))=12-(0+2)=10d_{21}=c_{21}-\left(u_2+v_1\right)=12-(0+2)=10d21=c21−(u2+v1)=12−(0+2)=10
  6. d 32 = c 32 − ( u 3 + v 2 ) = 14 − ( − 2 + 8 ) = 8 d 32 = c 32 − u 3 + v 2 = 14 − ( − 2 + 8 ) = 8 d_(32)=c_(32)-(u_(3)+v_(2))=14-(-2+8)=8d_{32}=c_{32}-\left(u_3+v_2\right)=14-(-2+8)=8d32=c32−(u3+v2)=14−(−2+8)=8
  7. d 33 = c 33 − ( u 3 + v 3 ) = 16 − ( − 2 + 9 ) = 9 d 33 = c 33 − u 3 + v 3 = 16 − ( − 2 + 9 ) = 9 d_(33)=c_(33)-(u_(3)+v_(3))=16-(-2+9)=9d_{33}=c_{33}-\left(u_3+v_3\right)=16-(-2+9)=9d33=c33−(u3+v3)=16−(−2+9)=9
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4Supply u i u i u_(i)u_iui
S 1 S 1 S_(1)S_1S1 10 [ 16 ] 10 [ 16 ] 10[16]10[16]10[16] 0 ( 15 ) 0 ( 15 ) 0(15)0(15)0(15) 20 [ 19 ] 20 [ 19 ] 20[19]20[19]20[19] 11 [ − 1 ] 11 [ − 1 ] 11[-1]11[-1]11[−1]15 u 1 = − 8 u 1 = − 8 u_(1)=-8u_1=-8u1=−8
S 2 S 2 S_(2)S_2S2 12 [ 10 ] 12 [ 10 ] 12[10]12[10]12[10] 8 ( 5 ) 8 ( 5 ) 8(5)8(5)8(5) 9 ( 15 ) 9 ( 15 ) 9(15)9(15)9(15) 20 ( 5 ) 20 ( 5 ) 20(5)20(5)20(5)25 u 2 = 0 u 2 = 0 u_(2)=0u_2=0u2=0
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5) 14 [ 8 ] 14 [ 8 ] 14[8]14[8]14[8] 16 [ 9 ] 16 [ 9 ] 16[9]16[9]16[9] 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10 u 3 = − 2 u 3 = − 2 u_(3)=-2u_3=-2u3=−2
Demand5201510
v j v j v_(j)v_jvj v 1 = 2 v 1 = 2 v_(1)=2v_1=2v1=2 v 2 = 8 v 2 = 8 v_(2)=8v_2=8v2=8 v 3 = 9 v 3 = 9 v_(3)=9v_3=9v3=9 v 4 = 20 v 4 = 20 v_(4)=20v_4=20v4=20
D_(1) D_(2) D_(3) D_(4) Supply u_(i) S_(1) 10[16] 0(15) 20[19] 11[-1] 15 u_(1)=-8 S_(2) 12[10] 8(5) 9(15) 20(5) 25 u_(2)=0 S_(3) 0(5) 14[8] 16[9] 18(5) 10 u_(3)=-2 Demand 5 20 15 10 v_(j) v_(1)=2 v_(2)=8 v_(3)=9 v_(4)=20 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | $u_i$ | | :— | :— | :— | :— | :— | :— | :— | | $S_1$ | $10[16]$ | $0(15)$ | $20[19]$ | $11[-1]$ | 15 | $u_1=-8$ | | $S_2$ | $12[10]$ | $8(5)$ | $9(15)$ | $20(5)$ | 25 | $u_2=0$ | | $S_3$ | $0(5)$ | $14[8]$ | $16[9]$ | $18(5)$ | 10 | $u_3=-2$ | | Demand | 5 | 20 | 15 | 10 | | | | $v_j$ | $v_1=2$ | $v_2=8$ | $v_3=9$ | $v_4=20$ | | |
  1. Now choose the minimum negative value from all d i j d i j d_(ij)d_{i j}dij (opportunity cost) = d 14 = [ − 1 ] = d 14 = [ − 1 ] =d_(14)=[-1]=d_{14}=[-1]=d14=[−1] and draw a closed path from S 1 D 4 S 1 D 4 S_(1)D_(4)S_1 D_4S1D4.
    Closed path is S 1 D 4 → S 1 D 2 → S 2 D 2 → S 2 D 4 S 1 D 4 → S 1 D 2 → S 2 D 2 → S 2 D 4 S_(1)D_(4)rarrS_(1)D_(2)rarrS_(2)D_(2)rarrS_(2)D_(4)S_1 D_4 \rightarrow S_1 D_2 \rightarrow S_2 D_2 \rightarrow S_2 D_4S1D4→S1D2→S2D2→S2D4
    Closed path and plus/minus sign allocation…
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4Supply u i u i u_(i)u_iui
S 1 S 1 S_(1)S_1S1 10 [ 16 ] 10 [ 16 ] 10[16]10[16]10[16] 0 ( 15 ) ( − ) 0 ( 15 ) ( − ) 0(15)(-)0(15)(-)0(15)(−) 20 [ 19 ] 20 [ 19 ] 20[19]20[19]20[19] 11 [ − 1 ] ( + ) 11 [ − 1 ] ( + ) 11[-1](+)11[-1](+)11[−1](+)15 u 1 = − 8 u 1 = − 8 u_(1)=-8u_1=-8u1=−8
S 2 S 2 S_(2)S_2S2 12 [ 10 ] 12 [ 10 ] 12[10]12[10]12[10] 8 ( 5 ) ( + ) 8 ( 5 ) ( + ) 8(5)(+)8(5)(+)8(5)(+) 9 ( 15 ) 9 ( 15 ) 9(15)9(15)9(15) 20 ( 5 ) ( − ) 20 ( 5 ) ( − ) 20(5)(-)20(5)(-)20(5)(−)25 u 2 = 0 u 2 = 0 u_(2)=0u_2=0u2=0
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5) 14 [ 8 ] 14 [ 8 ] 14[8]14[8]14[8] 16 [ 9 ] 16 [ 9 ] 16[9]16[9]16[9] 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10 u 3 = − 2 u 3 = − 2 u_(3)=-2u_3=-2u3=−2
Demand5201510
v j v j v_(j)v_jvj v 1 = 2 v 1 = 2 v_(1)=2v_1=2v1=2 v 2 = 8 v 2 = 8 v_(2)=8v_2=8v2=8 v 3 = 9 v 3 = 9 v_(3)=9v_3=9v3=9 v 4 = 20 v 4 = 20 v_(4)=20v_4=20v4=20
D_(1) D_(2) D_(3) D_(4) Supply u_(i) S_(1) 10[16] 0(15)(-) 20[19] 11[-1](+) 15 u_(1)=-8 S_(2) 12[10] 8(5)(+) 9(15) 20(5)(-) 25 u_(2)=0 S_(3) 0(5) 14[8] 16[9] 18(5) 10 u_(3)=-2 Demand 5 20 15 10 v_(j) v_(1)=2 v_(2)=8 v_(3)=9 v_(4)=20 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | $u_i$ | | :— | :— | :— | :— | :— | :— | :— | | $S_1$ | $10[16]$ | $0(15)(-)$ | $20[19]$ | $11[-1](+)$ | 15 | $u_1=-8$ | | $S_2$ | $12[10]$ | $8(5)(+)$ | $9(15)$ | $20(5)(-)$ | 25 | $u_2=0$ | | $S_3$ | $0(5)$ | $14[8]$ | $16[9]$ | $18(5)$ | 10 | $u_3=-2$ | | Demand | 5 | 20 | 15 | 10 | | | | $v_j$ | $v_1=2$ | $v_2=8$ | $v_3=9$ | $v_4=20$ | | |
  1. Minimum allocated value among all negative position (-) on closed path = 5 = 5 =5=5=5 Substract 5 from all (-) and Add it to all (+)
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4Supply
S 1 S 1 S_(1)S_1S110 0 ( 10 ) 0 ( 10 ) 0(10)0(10)0(10)20 11 ( 5 ) 11 ( 5 ) 11(5)11(5)11(5)15
S 2 S 2 S_(2)S_2S212 8 ( 10 ) 8 ( 10 ) 8(10)8(10)8(10) 9 ( 15 ) 9 ( 15 ) 9(15)9(15)9(15)2025
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5)1416 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10
Demand5201510
D_(1) D_(2) D_(3) D_(4) Supply S_(1) 10 0(10) 20 11(5) 15 S_(2) 12 8(10) 9(15) 20 25 S_(3) 0(5) 14 16 18(5) 10 Demand 5 20 15 10 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | | :— | :— | :— | :— | :— | :— | | $S_1$ | 10 | $0(10)$ | 20 | $11(5)$ | 15 | | $S_2$ | 12 | $8(10)$ | $9(15)$ | 20 | 25 | | $S_3$ | $0(5)$ | 14 | 16 | $18(5)$ | 10 | | Demand | 5 | 20 | 15 | 10 | |
  1. Repeat the step 1 to 4 , until an optimal solution is obtained.
Iteration-2 of optimality test
Find u i u i u_(i)u_iui and v j v j v_(j)v_jvj for all occupied cells ( i , j ) ( i , j ) (i,j)(\mathrm{i}, \mathrm{j})(i,j), where c i j = u i + v j c i j = u i + v j c_(ij)=u_(i)+v_(j)c_{i j}=u_i+v_jcij=ui+vj
  1. Substituting, u 1 = 0 u 1 = 0 u_(1)=0u_1=0u1=0, we get
  2. c 12 = u 1 + v 2 ⇒ v 2 = c 12 − u 1 ⇒ v 2 = 0 − 0 ⇒ v 2 = 0 c 12 = u 1 + v 2 ⇒ v 2 = c 12 − u 1 ⇒ v 2 = 0 − 0 ⇒ v 2 = 0 c_(12)=u_(1)+v_(2)=>v_(2)=c_(12)-u_(1)=>v_(2)=0-0=>v_(2)=0c_{12}=u_1+v_2 \Rightarrow v_2=c_{12}-u_1 \Rightarrow v_2=0-0 \Rightarrow v_2=0c12=u1+v2⇒v2=c12−u1⇒v2=0−0⇒v2=0
  3. c 22 = u 2 + v 2 ⇒ u 2 = c 22 − v 2 ⇒ u 2 = 8 − 0 ⇒ u 2 = 8 c 22 = u 2 + v 2 ⇒ u 2 = c 22 − v 2 ⇒ u 2 = 8 − 0 ⇒ u 2 = 8 c_(22)=u_(2)+v_(2)=>u_(2)=c_(22)-v_(2)=>u_(2)=8-0=>u_(2)=8c_{22}=u_2+v_2 \Rightarrow u_2=c_{22}-v_2 \Rightarrow u_2=8-0 \Rightarrow u_2=8c22=u2+v2⇒u2=c22−v2⇒u2=8−0⇒u2=8
  4. c 23 = u 2 + v 3 ⇒ v 3 = c 23 − u 2 ⇒ v 3 = 9 − 8 ⇒ v 3 = 1 c 23 = u 2 + v 3 ⇒ v 3 = c 23 − u 2 ⇒ v 3 = 9 − 8 ⇒ v 3 = 1 c_(23)=u_(2)+v_(3)=>v_(3)=c_(23)-u_(2)=>v_(3)=9-8=>v_(3)=1c_{23}=u_2+v_3 \Rightarrow v_3=c_{23}-u_2 \Rightarrow v_3=9-8 \Rightarrow v_3=1c23=u2+v3⇒v3=c23−u2⇒v3=9−8⇒v3=1
  5. c 14 = u 1 + v 4 ⇒ v 4 = c 14 − u 1 ⇒ v 4 = 11 − 0 ⇒ v 4 = 11 c 14 = u 1 + v 4 ⇒ v 4 = c 14 − u 1 ⇒ v 4 = 11 − 0 ⇒ v 4 = 11 c_(14)=u_(1)+v_(4)=>v_(4)=c_(14)-u_(1)=>v_(4)=11-0=>v_(4)=11c_{14}=u_1+v_4 \Rightarrow v_4=c_{14}-u_1 \Rightarrow v_4=11-0 \Rightarrow v_4=11c14=u1+v4⇒v4=c14−u1⇒v4=11−0⇒v4=11
  6. c 34 = u 3 + v 4 ⇒ u 3 = c 34 − v 4 ⇒ u 3 = 18 − 11 ⇒ u 3 = 7 c 34 = u 3 + v 4 ⇒ u 3 = c 34 − v 4 ⇒ u 3 = 18 − 11 ⇒ u 3 = 7 c_(34)=u_(3)+v_(4)=>u_(3)=c_(34)-v_(4)=>u_(3)=18-11=>u_(3)=7c_{34}=u_3+v_4 \Rightarrow u_3=c_{34}-v_4 \Rightarrow u_3=18-11 \Rightarrow u_3=7c34=u3+v4⇒u3=c34−v4⇒u3=18−11⇒u3=7
  7. c 31 = u 3 + v 1 ⇒ v 1 = c 31 − u 3 ⇒ v 1 = 0 − 7 ⇒ v 1 = − 7 c 31 = u 3 + v 1 ⇒ v 1 = c 31 − u 3 ⇒ v 1 = 0 − 7 ⇒ v 1 = − 7 c_(31)=u_(3)+v_(1)=>v_(1)=c_(31)-u_(3)=>v_(1)=0-7=>v_(1)=-7c_{31}=u_3+v_1 \Rightarrow v_1=c_{31}-u_3 \Rightarrow v_1=0-7 \Rightarrow v_1=-7c31=u3+v1⇒v1=c31−u3⇒v1=0−7⇒v1=−7
D 1 D 2 D 3 D 4 Supply u i S 1 10 0 ( 10 ) 20 11 ( 5 ) 15 u 1 = 0 S 2 12 8 ( 10 ) 9 ( 15 ) 20 25 u 2 = 8 S 3 0 ( 5 ) 14 16 18 ( 5 ) 10 u 3 = 7 Demand 5 20 15 10 v j v 1 = − 7 v 2 = 0 v 3 = 1 v 4 = 11 D 1 D 2 D 3 D 4  Supply  u i S 1 10 0 ( 10 ) 20 11 ( 5 ) 15 u 1 = 0 S 2 12 8 ( 10 ) 9 ( 15 ) 20 25 u 2 = 8 S 3 0 ( 5 ) 14 16 18 ( 5 ) 10 u 3 = 7  Demand  5 20 15 10 v j v 1 = − 7 v 2 = 0 v 3 = 1 v 4 = 11 {:[,D_(1),D_(2),D_(3),D_(4),” Supply “,u_(i)],[S_(1),10,0(10),20,11(5),15,u_(1)=0],[S_(2),12,8(10),9(15),20,25,u_(2)=8],[S_(3),0(5),14,16,18(5),10,u_(3)=7],[” Demand “,5,20,15,10,,],[v_(j),v_(1)=-7,v_(2)=0,v_(3)=1,v_(4)=11,,]:}\begin{array}{|l|l|l|l|l||l|l|} \hline & D_1 & D_2 & D_3 & D_4 & \text { Supply } & u_i \\ \hline S_1 & 10 & 0(10) & 20 & 11(5) & 15 & u_1=0 \\ \hline S_2 & 12 & 8(10) & 9(15) & 20 & 25 & u_2=8 \\ \hline S_3 & 0(5) & 14 & 16 & 18(5) & 10 & u_3=7 \\ \hline \hline \text { Demand } & 5 & 20 & 15 & 10 & & \\ \hline v_j & v_1=-7 & v_2=0 & v_3=1 & v_4=11 & & \\ \hline \end{array}D1D2D3D4 Supply uiS1100(10)2011(5)15u1=0S2128(10)9(15)2025u2=8S30(5)141618(5)10u3=7 Demand 5201510vjv1=−7v2=0v3=1v4=11
  1. Find d i j d i j d_(ij)d_{i j}dij for all unoccupied cells ( i , j ) ( i , j ) (i,j)(\mathrm{i}, \mathrm{j})(i,j), where d i j = c i j − ( u i + v j ) d i j = c i j − u i + v j d_(ij)=c_(ij)-(u_(i)+v_(j))d_{i j}=c_{i j}-\left(u_i+v_j\right)dij=cij−(ui+vj)
  1. d 11 = c 11 − ( u 1 + v 1 ) = 10 − ( 0 − 7 ) = 17 d 11 = c 11 − u 1 + v 1 = 10 − ( 0 − 7 ) = 17 d_(11)=c_(11)-(u_(1)+v_(1))=10-(0-7)=17d_{11}=c_{11}-\left(u_1+v_1\right)=10-(0-7)=17d11=c11−(u1+v1)=10−(0−7)=17
  2. d 13 = c 13 − ( u 1 + v 3 ) = 20 − ( 0 + 1 ) = 19 d 13 = c 13 − u 1 + v 3 = 20 − ( 0 + 1 ) = 19 d_(13)=c_(13)-(u_(1)+v_(3))=20-(0+1)=19d_{13}=c_{13}-\left(u_1+v_3\right)=20-(0+1)=19d13=c13−(u1+v3)=20−(0+1)=19
  3. d 21 = c 21 − ( u 2 + v 1 ) = 12 − ( 8 − 7 ) = 11 d 21 = c 21 − u 2 + v 1 = 12 − ( 8 − 7 ) = 11 d_(21)=c_(21)-(u_(2)+v_(1))=12-(8-7)=11d_{21}=c_{21}-\left(u_2+v_1\right)=12-(8-7)=11d21=c21−(u2+v1)=12−(8−7)=11
  4. d 24 = c 24 − ( u 2 + v 4 ) = 20 − ( 8 + 11 ) = 1 d 24 = c 24 − u 2 + v 4 = 20 − ( 8 + 11 ) = 1 d_(24)=c_(24)-(u_(2)+v_(4))=20-(8+11)=1d_{24}=c_{24}-\left(u_2+v_4\right)=20-(8+11)=1d24=c24−(u2+v4)=20−(8+11)=1
  5. d 32 = c 32 − ( u 3 + v 2 ) = 14 − ( 7 + 0 ) = 7 d 32 = c 32 − u 3 + v 2 = 14 − ( 7 + 0 ) = 7 d_(32)=c_(32)-(u_(3)+v_(2))=14-(7+0)=7d_{32}=c_{32}-\left(u_3+v_2\right)=14-(7+0)=7d32=c32−(u3+v2)=14−(7+0)=7
  6. d 33 = c 33 − ( u 3 + v 3 ) = 16 − ( 7 + 1 ) = 8 d 33 = c 33 − u 3 + v 3 = 16 − ( 7 + 1 ) = 8 d_(33)=c_(33)-(u_(3)+v_(3))=16-(7+1)=8d_{33}=c_{33}-\left(u_3+v_3\right)=16-(7+1)=8d33=c33−(u3+v3)=16−(7+1)=8
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4Supply u i u i u_(i)u_iui
S 1 S 1 S_(1)S_1S1 10 [ 17 ] 10 [ 17 ] 10[17]10[17]10[17] 0 ( 10 ) 0 ( 10 ) 0(10)0(10)0(10) 20 [ 19 ] 20 [ 19 ] 20[19]20[19]20[19] 11 ( 5 ) 11 ( 5 ) 11(5)11(5)11(5)15 u 1 = 0 u 1 = 0 u_(1)=0u_1=0u1=0
S 2 S 2 S_(2)S_2S2 12 [ 11 ] 12 [ 11 ] 12[11]12[11]12[11] 8 ( 10 ) 8 ( 10 ) 8(10)8(10)8(10) 9 ( 15 ) 9 ( 15 ) 9(15)9(15)9(15) 20 [ 1 ] 20 [ 1 ] 20[1]20[1]20[1]25 u 2 = 8 u 2 = 8 u_(2)=8u_2=8u2=8
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5) 14 [ 7 ] 14 [ 7 ] 14[7]14[7]14[7] 16 [ 8 ] 16 [ 8 ] 16[8]16[8]16[8] 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10 u 3 = 7 u 3 = 7 u_(3)=7u_3=7u3=7
Demand5201510
v j v j v_(j)v_jvj v 1 = − 7 v 1 = − 7 v_(1)=-7v_1=-7v1=−7 v 2 = 0 v 2 = 0 v_(2)=0v_2=0v2=0 v 3 = 1 v 3 = 1 v_(3)=1v_3=1v3=1 v 4 = 11 v 4 = 11 v_(4)=11v_4=11v4=11
D_(1) D_(2) D_(3) D_(4) Supply u_(i) S_(1) 10[17] 0(10) 20[19] 11(5) 15 u_(1)=0 S_(2) 12[11] 8(10) 9(15) 20[1] 25 u_(2)=8 S_(3) 0(5) 14[7] 16[8] 18(5) 10 u_(3)=7 Demand 5 20 15 10 v_(j) v_(1)=-7 v_(2)=0 v_(3)=1 v_(4)=11 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | $u_i$ | | :— | :— | :— | :— | :— | :— | :— | | $S_1$ | $10[17]$ | $0(10)$ | $20[19]$ | $11(5)$ | 15 | $u_1=0$ | | $S_2$ | $12[11]$ | $8(10)$ | $9(15)$ | $20[1]$ | 25 | $u_2=8$ | | $S_3$ | $0(5)$ | $14[7]$ | $16[8]$ | $18(5)$ | 10 | $u_3=7$ | | Demand | 5 | 20 | 15 | 10 | | | | $v_j$ | $v_1=-7$ | $v_2=0$ | $v_3=1$ | $v_4=11$ | | |
Since all d i j ≥ 0 d i j ≥ 0 d_(ij) >= 0d_{i j} \geq 0dij≥0.
So final optimal solution is arrived.
D 1 D 1 D_(1)D_1D1 D 2 D 2 D_(2)D_2D2 D 3 D 3 D_(3)D_3D3 D 4 D 4 D_(4)D_4D4Supply
S 1 S 1 S_(1)S_1S110 0 ( 1 0 ) 0 ( 1 0 ) 0(10)0(\mathbf{1 0 )}0(10)20 11 ( 5 ) 11 ( 5 ) 11(5)11(5)11(5)15
S 2 S 2 S_(2)S_2S212 8 ( 10 ) 8 ( 10 ) 8(10)8(10)8(10) 9 ( 1 5 ) 9 ( 1 5 ) 9(15)9(\mathbf{1 5 )}9(15)2025
S 3 S 3 S_(3)S_3S3 0 ( 5 ) 0 ( 5 ) 0(5)0(5)0(5)1416 18 ( 5 ) 18 ( 5 ) 18(5)18(5)18(5)10
Demand5201510
D_(1) D_(2) D_(3) D_(4) Supply S_(1) 10 0(10) 20 11(5) 15 S_(2) 12 8(10) 9(15) 20 25 S_(3) 0(5) 14 16 18(5) 10 Demand 5 20 15 10 | | $D_1$ | $D_2$ | $D_3$ | $D_4$ | Supply | | :— | :— | :— | :— | :— | :— | | $S_1$ | 10 | $0(\mathbf{1 0 )}$ | 20 | $11(5)$ | 15 | | $S_2$ | 12 | $8(10)$ | $9(\mathbf{1 5 )}$ | 20 | 25 | | $S_3$ | $0(5)$ | 14 | 16 | $18(5)$ | 10 | | Demand | 5 | 20 | 15 | 10 | |
The minimum total transportation cost = 0 × 10 + 11 × 5 + 8 × 10 + 9 × 15 + 0 × 5 + 18 × 5 = 360  The minimum total transportation cost  = 0 × 10 + 11 × 5 + 8 × 10 + 9 × 15 + 0 × 5 + 18 × 5 = 360 ” The minimum total transportation cost “=0xx10+11 xx5+8xx10+9xx15+0xx5+18 xx5=360\text { The minimum total transportation cost }=0 \times 10+11 \times 5+8 \times 10+9 \times 15+0 \times 5+18 \times 5=360 The minimum total transportation cost =0×10+11×5+8×10+9×15+0×5+18×5=360
खण्ड ‘B’ SECTION ‘B’
5.(a) z = y f ( x ) + x g ( y ) z = y f ( x ) + x g ( y ) z=yf(x)+xg(y)z=y f(x)+x g(y)z=yf(x)+xg(y) से स्वैच्छिक फलनों f ( x ) f ( x ) f(x)f(x)f(x) व g ( y ) g ( y ) g(y)g(y)g(y) का विलोपन कर आंशिक अवकल समीकरण बनाइए तथा इसकी प्रकृति (दीर्घवृत्तीय, अतिपरवलीय या परवलीय) x > 0 , y > 0 x > 0 , y > 0 x > 0,y > 0x>0, y>0x>0,y>0 क्षेत्र में इंगित कीजिए ।
Form a partial differential equation by eliminating the arbitrary functions f ( x ) f ( x ) f(x)f(x)f(x) and g ( y ) g ( y ) g(y)g(y)g(y) from z = y f ( x ) + x g ( y ) z = y f ( x ) + x g ( y ) z=yf(x)+xg(y)z=y f(x)+x g(y)z=yf(x)+xg(y) and specify its nature (elliptic, hyperbolic or parabolic) in the region x > 0 , y > 0 x > 0 , y > 0 x > 0,y > 0x>0, y>0x>0,y>0.
Answer:
Introduction:
The problem requires forming a partial differential equation by eliminating the arbitrary functions f ( x ) f ( x ) f(x)f(x)f(x) and g ( y ) g ( y ) g(y)g(y)g(y) from the given equation z = y f ( x ) + x g ( y ) z = y f ( x ) + x g ( y ) z=yf(x)+xg(y)z=yf(x)+xg(y)z=yf(x)+xg(y). Additionally, it asks for specifying the nature of this equation in the region x > 0 , y > 0 x > 0 , y > 0 x > 0,y > 0x>0, y>0x>0,y>0.
Solution:
Step 1: Given Equation
We are given the equation:
z = y f ( x ) + x g ( y ) ( 1 ) z = y f ( x ) + x g ( y ) ( 1 ) z=yf(x)+xg(y)(1)z = yf(x) + xg(y)\ \ \ \ \ \ \ \ \ (1)z=yf(x)+xg(y)         (1)
Step 2: Partial Differentiation
We differentiate equation (1) partially with respect to x x xxx and y y yyy:
∂ z ∂ x = y f ′ ( x ) + g ( y ) ( 2 ) ∂ z ∂ y = f ( x ) + x g ′ ( y ) ( 3 ) ∂ z ∂ x = y f ′ ( x ) + g ( y ) ( 2 ) ∂ z ∂ y = f ( x ) + x g ′ ( y ) ( 3 ) {:[(del z)/(del x)=yf^(‘)(x)+g(y)(2)],[(del z)/(del y)=f(x)+xg^(‘)(y)(3)]:}\begin{aligned} &\frac{\partial z}{\partial x} = yf'(x) + g(y)\ \ \ \ \ \ \ \ \ (2) \\ &\frac{\partial z}{\partial y} = f(x) + xg'(y)\ \ \ \ \ \ \ \ \ (3) \end{aligned}∂z∂x=yf′(x)+g(y)         (2)∂z∂y=f(x)+xg′(y)         (3)
Step 3: Second Partial Differentiation
We differentiate equation (3) with respect to x x xxx:
∂ 2 z ∂ x ∂ y = f ′ ( x ) + g ′ ( y ) ( 4 ) ∂ 2 z ∂ x ∂ y = f ′ ( x ) + g ′ ( y ) ( 4 ) (del^(2)z)/(del x del y)=f^(‘)(x)+g^(‘)(y)(4)\frac{\partial^2 z}{\partial x \partial y} = f'(x) + g'(y)\ \ \ \ \ \ \ \ \ (4)∂2z∂x∂y=f′(x)+g′(y)         (4)
Step 4: Expressing f ′ ( x ) f ′ ( x ) f^(‘)(x)f'(x)f′(x) and g ′ ( y ) g ′ ( y ) g^(‘)(y)g'(y)g′(y)
From equations (2) and (3), we express f ′ ( x ) f ′ ( x ) f^(‘)(x)f'(x)f′(x) and g ′ ( y ) g ′ ( y ) g^(‘)(y)g'(y)g′(y) as follows:
f ′ ( x ) = 1 y ( ∂ z ∂ x − g ( y ) ) g ′ ( y ) = 1 x ( ∂ z ∂ y − f ( x ) ) f ′ ( x ) = 1 y ∂ z ∂ x − g ( y ) g ′ ( y ) = 1 x ∂ z ∂ y − f ( x ) {:[f^(‘)(x)=(1)/(y)((del z)/(del x)-g(y))],[g^(‘)(y)=(1)/(x)((del z)/(del y)-f(x))]:}\begin{aligned} f'(x) &= \frac{1}{y}\left(\frac{\partial z}{\partial x} – g(y)\right) \\ g'(y) &= \frac{1}{x}\left(\frac{\partial z}{\partial y} – f(x)\right) \end{aligned}f′(x)=1y(∂z∂x−g(y))g′(y)=1x(∂z∂y−f(x))
Step 5: Substituting into Equation (4)
Substituting the expressions for f ′ ( x ) f ′ ( x ) f^(‘)(x)f'(x)f′(x) and g ′ ( y ) g ′ ( y ) g^(‘)(y)g'(y)g′(y) into equation (4), we obtain:
∂ 2 z ∂ x ∂ y = 1 y ( ∂ z ∂ x − g ( y ) ) + 1 x ( ∂ z ∂ y − f ( x ) ) ⇒ x y ∂ 2 z ∂ x ∂ y = x ∂ z ∂ x + y ∂ z ∂ y − { x g ( y ) + y f ( x ) } ⇒ x y ∂ 2 z ∂ x ∂ y = x ∂ z ∂ x + y ∂ z ∂ y − z x y ∂ 2 z ∂ x ∂ y − x ∂ z ∂ x − y ∂ z ∂ y + z = 0 ∂ 2 z ∂ x ∂ y = 1 y ∂ z ∂ x − g ( y ) + 1 x ∂ z ∂ y − f ( x ) ⇒ x y ∂ 2 z ∂ x ∂ y = x ∂ z ∂ x + y ∂ z ∂ y − { x g ( y ) + y f ( x ) } ⇒ x y ∂ 2 z ∂ x ∂ y = x ∂ z ∂ x + y ∂ z ∂ y − z x y ∂ 2 z ∂ x ∂ y − x ∂ z ∂ x − y ∂ z ∂ y + z = 0 {:[(del^(2)z)/(del x del y)=(1)/(y)((del z)/(del x)-g(y))+(1)/(x)((del z)/(del y)-f(x))],[=>xy(del^(2)z)/(del x del y)=x(del z)/(del x)+y(del z)/(del y)-{xg(y)+yf(x)}],[=>xy(del^(2)z)/(del x del y)=x(del z)/(del x)+y(del z)/(del y)-z],[xy(del^(2)z)/(del x del y)-x(del z)/(del x)-y(del z)/(del y)+z=0]:}\begin{aligned} &\frac{\partial^2 z}{\partial x \partial y} = \frac{1}{y}\left(\frac{\partial z}{\partial x} – g(y)\right) + \frac{1}{x}\left(\frac{\partial z}{\partial y} – f(x)\right) \\ &\Rightarrow x y \frac{\partial^2 z}{\partial x \partial y} = x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} – \{xg(y) + yf(x)\} \\ &\Rightarrow x y \frac{\partial^2 z}{\partial x \partial y} = x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} – z \\ & x y \frac{\partial^2 z}{\partial x \partial y} – x \frac{\partial z}{\partial x} – y \frac{\partial z}{\partial y} + z = 0 \end{aligned}∂2z∂x∂y=1y(∂z∂x−g(y))+1x(∂z∂y−f(x))⇒xy∂2z∂x∂y=x∂z∂x+y∂z∂y−{xg(y)+yf(x)}⇒xy∂2z∂x∂y=x∂z∂x+y∂z∂y−zxy∂2z∂x∂y−x∂z∂x−y∂z∂y+z=0
Step 6: Compare with General Form
Comparing the obtained equation with the general form R r + S s + T t + f ( x , y , z ) = 0 R r + S s + T t + f ( x , y , z ) = 0 R_(r)+S_(s)+T_(t)+f(x,y,z)=0R_r + S_s + T_t + f(x, y, z) = 0Rr+Ss+Tt+f(x,y,z)=0, we have:
R = 0 S = x y T = 0 R = 0 S = x y T = 0 {:[R=0],[S=xy],[T=0]:}\begin{aligned} R &= 0 \\ S &= xy \\ T &= 0 \end{aligned}R=0S=xyT=0
Step 7: Nature of the Equation
The discriminant of the equation is given by S 2 − 4 R T = x 2 y 2 > 0 S 2 − 4 R T = x 2 y 2 > 0 S^(2)-4RT=x^(2)y^(2) > 0S^2 – 4RT = x^2y^2 > 0S2−4RT=x2y2>0, indicating that it is a hyperbolic partial differential equation.
Conclusion:
The partial differential equation obtained from eliminating the arbitrary functions f ( x ) f ( x ) f(x)f(x)f(x) and g ( y ) g ( y ) g(y)g(y)g(y) from z = y f ( x ) + x g ( y ) z = y f ( x ) + x g ( y ) z=yf(x)+xg(y)z=yf(x)+xg(y)z=yf(x)+xg(y) is hyperbolic in the region x > 0 , y > 0 x > 0 , y > 0 x > 0,y > 0x>0, y>0x>0,y>0.
5.(b) दर्शाइए कि समीकरण : f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f(x)=cos((pi(x+1))/(8))+0*148 x-0*9062=0f(x)=\cos \frac{\pi(x+1)}{8}+0 \cdot 148 x-0 \cdot 9062=0f(x)=cos⁡π(x+1)8+0⋅148x−0⋅9062=0
का एक मूल अन्तराल ( − 1 , 0 ) ( − 1 , 0 ) (-1,0)(-1,0)(−1,0) में तथा एक मूल ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1) में है । ऋणात्मक मूल की न्यूटन-रॉफसन विधि से दशमलव के चार स्थान तक सही गणना कीजिए।
Show that the equation: f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0 ⋅ 148 x − 0 ⋅ 9062 = 0 f(x)=cos((pi(x+1))/(8))+0*148 x-0*9062=0f(x)=\cos \frac{\pi(x+1)}{8}+0 \cdot 148 x-0 \cdot 9062=0f(x)=cos⁡π(x+1)8+0⋅148x−0⋅9062=0
has one root in the interval ( − 1 , 0 ) ( − 1 , 0 ) (-1,0)(-1,0)(−1,0) and one in ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)(0,1). Calculate the negative root correct to four decimal places using Newton-Raphson method.
Answer:
Introduction:
The problem involves showing that the equation f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0.148 x − 0.9062 = 0 f ( x ) = cos ⁡ π ( x + 1 ) 8 + 0.148 x − 0.9062 = 0 f(x)=cos((pi(x+1))/(8))+0.148 x-0.9062=0f(x) = \cos \frac{\pi(x+1)}{8} + 0.148x – 0.9062 = 0