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Question:-1(e)

UPSC Mathematics (Optional) Previous Year Paper Solution | Paper-02 | IAS Mathematics 2018 Paper-02 Question-01 Question:-1(e)
\(\operatorname{cosec}^2 \theta=1+\cot ^2 \theta\)
An agricultural firm has 180 tons of nitrogen fertilizer, 250 tons of phosphate and 220 tons of potash. It will be able to sell a mixture of these substances in their respective ratio \(3: 3: 4\) at a profit of Rs. 1500 per ton and a mixture in the ratio \(2: 4: 2\) at a profit of Rs. 1200 per ton. Pose a linear programming problem to show how many tons of these two mixtures should be prepared to obtain the maximum profit.
Expert Answer
untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f
To formulate a Linear Programming (LP) problem, we need to define the decision variables, the objective function, and the constraints.

Decision Variables

Let:
  • x x xxx be the number of tons of the mixture in the ratio 3 : 3 : 4 3 : 3 : 4 3:3:43:3:43:3:4 to be prepared.
  • y y yyy be the number of tons of the mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22:4:22:4:2 to be prepared.

Objective Function

The firm wants to maximize its profit. The profit for the mixture in the ratio 3 : 3 : 4 3 : 3 : 4 3:3:43:3:43:3:4 is Rs. 1500 per ton, and for the mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22:4:22:4:2, it is Rs. 1200 per ton. So, the objective function to maximize is:
P = 1500 x + 1200 y P = 1500 x + 1200 y P=1500 x+1200 yP = 1500x + 1200yP=1500x+1200y

Constraints

The firm has limited amounts of nitrogen fertilizer, phosphate, and potash, so we need to set up constraints based on the availability of these substances.
Certainly! Here is the corrected version with x x xxx and y y yyy replaced by x 1 x 1 x_(1)x_1x1 and x 2 x 2 x_(2)x_2x2:

Nitrogen Fertilizer Constraint

For the mixture in the ratio 3 : 3 : 4 3 : 3 : 4 3:3:43:3:43:3:4, 3 parts out of 10 are nitrogen fertilizer. So, for every ton of this mixture, 3 10 3 10 (3)/(10)\frac{3}{10}310 tons of nitrogen fertilizer are used. Similarly, for the mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22:4:22:4:2, 2 parts out of 8 are nitrogen fertilizer, i.e., 2 8 = 1 4 2 8 = 1 4 (2)/(8)=(1)/(4)\frac{2}{8} = \frac{1}{4}28=14 tons of nitrogen fertilizer per ton of mixture. The firm has 180 tons of nitrogen fertilizer available, so:
3 10 x 1 + 1 4 x 2 ≤ 180 3 10 x 1 + 1 4 x 2 ≤ 180 (3)/(10)x_(1)+(1)/(4)x_(2) <= 180\frac{3}{10}x_1 + \frac{1}{4}x_2 \leq 180310x1+14x2≤180
6 x 1 + 5 x 2 ≤ 3600 6 x 1 + 5 x 2 ≤ 3600 6x_(1)+5x_(2) <= 36006x_1 + 5x_2 \leq 36006x1+5x2≤3600

Phosphate Constraint

For the mixture in the ratio 3 : 3 : 4 3 : 3 : 4 3:3:43:3:43:3:4, 3 parts out of 10 are phosphate. So, for every ton of this mixture, 3 10 3 10 (3)/(10)\frac{3}{10}310 tons of phosphate are used. For the mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22:4:22:4:2, 4 parts out of 8 are phosphate, i.e., 4 8 = 1 2 4 8 = 1 2 (4)/(8)=(1)/(2)\frac{4}{8} = \frac{1}{2}48=12 tons of phosphate per ton of mixture. The firm has 250 tons of phosphate available, so:
3 10 x 1 + 1 2 x 2 ≤ 250 3 10 x 1 + 1 2 x 2 ≤ 250 (3)/(10)x_(1)+(1)/(2)x_(2) <= 250\frac{3}{10}x_1 + \frac{1}{2}x_2 \leq 250310x1+12x2≤250
3 x 1 + 5 x 2 ≤ 2500 3 x 1 + 5 x 2 ≤ 2500 3x_(1)+5x_(2) <= 25003x_1 + 5x_2 \leq 25003x1+5x2≤2500

Potash Constraint

For the mixture in the ratio 3 : 3 : 4 3 : 3 : 4 3:3:43:3:43:3:4, 4 parts out of 10 are potash. So, for every ton of this mixture, 4 10 4 10 (4)/(10)\frac{4}{10}410 tons of potash are used. For the mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22:4:22:4:2, 2 parts out of 8 are potash, i.e., 2 8 = 1 4 2 8 = 1 4 (2)/(8)=(1)/(4)\frac{2}{8} = \frac{1}{4}28=14 tons of potash per ton of mixture. The firm has 220 tons of potash available, so:
4 10 x 1 + 1 4 x 2 ≤ 220 4 10 x 1 + 1 4 x 2 ≤ 220 (4)/(10)x_(1)+(1)/(4)x_(2) <= 220\frac{4}{10}x_1 + \frac{1}{4}x_2 \leq 220410x1+14x2≤220
8 x 1 + 5 x 2 ≤ 4400 8 x 1 + 5 x 2 ≤ 4400 8x_(1)+5x_(2) <= 44008x_1 + 5x_2 \leq 44008x1+5x2≤4400

Non-negativity Constraint

The number of tons of the mixtures cannot be negative:
x 1 ≥ 0 x 1 ≥ 0 x_(1) >= 0x_1 \geq 0x1≥0
x 2 ≥ 0 x 2 ≥ 0 x_(2) >= 0x_2 \geq 0x2≥0

Linear Programming Problem

To summarize, the Linear Programming problem is as follows:

Maximize

Z = 1500 x 1 + 1200 x 2 Z = 1500 x 1 + 1200 x 2 Z=1500x_(1)+1200x_(2)Z = 1500x_1 + 1200x_2Z=1500x1+1200x2

Subject to

6 x 1 + 5 x 2 ≤ 3600 3 x 1 + 5 x 2 ≤ 2500 8 x 1 + 5 x 2 ≤ 4400 x 1 , x 2 ≥ 0 6 x 1 + 5 x 2 ≤ 3600 3 x 1 + 5 x 2 ≤ 2500 8 x 1 + 5 x 2 ≤ 4400 x 1 , x 2 ≥ 0 {:[6x_(1)+5x_(2) <= 3600],[3x_(1)+5x_(2) <= 2500],[8x_(1)+5x_(2) <= 4400],[x_(1)”,”x_(2) >= 0]:}\begin{aligned} & 6x_1 + 5x_2 \leq 3600 \\ & 3x_1 + 5x_2 \leq 2500 \\ & 8x_1 + 5x_2 \leq 4400 \\ & x_1, x_2 \geq 0 \end{aligned}6x1+5x2≤36003x1+5x2≤25008x1+5x2≤4400x1,x2≥0
Now, let’s solve this Linear Programming problem.
The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate
  1. As the constraint-1 is of type ‘ ≤ ≤ <=\leq≤ ‘ we should add slack variable S 1 S 1 S_(1)S_1S1
  2. As the constraint-2 is of type ‘ ≤ ≤ <=\leq≤ ‘ we should add slack variable S 2 S 2 S_(2)S_2S2
  3. As the constraint-3 is of type ‘ ≤ ≤ <=\leq≤ ‘ we should add slack variable S 3 S 3 S_(3)S_3S3
After introducing slack variables
Max ⁡ Z = 1500 x 1 + 1200 x 2 + 0 S 1 + 0 S 2 + 0 S 3 Max ⁡ Z = 1500 x 1 + 1200 x 2 + 0 S 1 + 0 S 2 + 0 S 3 Max Z=1500x_(1)+1200x_(2)+0S_(1)+0S_(2)+0S_(3)\operatorname{Max} Z=1500 x_1+1200 x_2+0 S_1+0 S_2+0 S_3Max⁡Z=1500x1+1200x2+0S1+0S2+0S3
subject to
6 x 1 + 5 x 2 + S 1 = 3600 3 x 1 + 5 x 2 + S 2 = 2500 8 x 1 + 5 x 2 + S 3 = 4400 6 x 1 + 5 x 2 + S 1 = 3600 3 x 1 + 5 x 2 + S 2 = 2500 8 x 1 + 5 x 2 + S 3 = 4400 {:[6x_(1)+5x_(2)+S_(1)=3600],[3x_(1)+5x_(2)+S_(2)=2500],[8x_(1)+5x_(2)+S_(3)=4400]:}\begin{aligned} 6 x_1+5 x_2+S_1 & =3600 \\ 3 x_1+5 x_2+S_2 & =2500 \\ 8 x_1+5 x_2+S_3 & =4400 \end{aligned}6x1+5x2+S1=36003x1+5x2+S2=25008x1+5x2+S3=4400
and x 1 , x 2 , S 1 , S 2 , S 3 ≥ 0 x 1 , x 2 , S 1 , S 2 , S 3 ≥ 0 x_(1),x_(2),S_(1),S_(2),S_(3) >= 0x_1, x_2, S_1, S_2, S_3 \geq 0x1,x2,S1,S2,S3≥0
Iteration-1 C j 1500 1200 0 0 0 B C B X B x 1 x 2 S 1 S 2 S 3 X B x 1 S 1 S 2 0 2500 3 5 0 1 0 2500 3 = 833.3333 S 3 0 4400 ( 8 ) 5 0 0 1 4400 8 = 550 → Z = 0 Z j 0 0 0 0 0 C j − Z j 1500 ↑ 1200 0 0 0  Iteration-1  C j 1500 1200 0 0 0 B C B X B x 1 x 2 S 1 S 2 S 3 X B x 1 S 1 S 2 0 2500 3 5 0 1 0 2500 3 = 833.3333 S 3 0 4400 ( 8 ) 5 0 0 1 4400 8 = 550 → Z = 0 Z j 0 0 0 0 0 C j − Z j 1500 ↑ 1200 0 0 0 {:[” Iteration-1 “,,C_(j),1500,1200,0,0,0,],[B,C_(B),X_(B),x_(1),x_(2),S_(1),S_(2),S_(3),{:[(X_(B))/(x_(1))],[S_(1)]:}],[S_(2),0,2500,3,5,0,1,0,(2500)/(3)=833.3333],[S_(3),0,4400,(8),5,0,0,1,(4400)/(8)=550 rarr],[Z=0,,Z_(j),0,0,0,0,0,],[,,C_(j)-Z_(j),1500 uarr,1200,0,0,0,]:}\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text { Iteration-1 } & & C_j & 1500 & 1200 & 0 & 0 & 0 & \\ \hline \boldsymbol{B} & C_{\boldsymbol{B}} & \boldsymbol{X}_{\boldsymbol{B}} & \boldsymbol{x}_1 & \boldsymbol{x}_2 & \boldsymbol{S}_{\mathbf{1}} & \boldsymbol{S}_{\mathbf{2}} & \boldsymbol{S}_{\mathbf{3}} & \begin{array}{c} \frac{\boldsymbol{X}_{\boldsymbol{B}}}{\boldsymbol{x}_{\mathbf{1}}} \\ \hline S_1 \end{array} \\ \hline S_2 & 0 & 2500 & 3 & 5 & 0 & 1 & 0 & \frac{2500}{3}=833.3333 \\ \hline S_3 & 0 & 4400 & (8) & 5 & 0 & 0 & 1 & \frac{4400}{8}=550 \rightarrow \\ \hline \boldsymbol{Z}=\mathbf{0} & & \boldsymbol{Z}_{\boldsymbol{j}} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \mathbf{0} & \\ \hline & & C_j-Z_j & 1500 \uparrow & 1200 & 0 & 0 & 0 & \\ \hline \end{array} Iteration-1 Cj15001200000BCBXBx1x2S1S2S3XBx1S1S2025003501025003=833.3333S304400(8)500144008=550→Z=0Zj00000Cj−Zj1500↑1200000
Positive maximum C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj is 1500 and its column index is 1 . So, the entering variable is x 1 x 1 x_(1)x_1x1.
Minimum ratio is 550 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 8 .
Entering = x 1 = x 1 =x_(1)=x_1=x1, Departing = S 3 = S 3 =S_(3)=S_3=S3, Key Element = 8 = 8 =8=8=8
R 3 ( R 3 ( R_(3)(R_3(R3( new ) = R 3 ( ) = R 3 ( )=R_(3)()=R_3()=R3( old ) ÷ 8 ) ÷ 8 )-:8) \div 8)÷8
R 1 ( R 1 ( R_(1)(R_1(R1( new ) = R 1 ( ) = R 1 ( )=R_(1)()=R_1()=R1( old ) − 6 R 3 ( ) − 6 R 3 ( )-6R_(3)()-6 R_3()−6R3( new ) ) )))
R 2 ( R 2 ( R_(2)(R_2(R2( new ) = R 2 ( ) = R 2 ( )=R_(2)()=R_2()=R2( old ) − 3 R 3 ( ) − 3 R 3 ( )-3R_(3)()-3 R_3()−3R3( new ) ) )))
Iteration-2 C j C j C_(j)C_jCj 1500 1200 0 0 0
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 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
s 1 s 1 s_(1)s_1s1 0 300 0 (1.25) 1 0 -0.75 300 1.25 = 240 → 300 1.25 = 240 → (300)/(1.25)=240 rarr\frac{300}{1.25}=240 \rightarrow3001.25=240→
S 2 S 2 S_(2)S_2S2 0 850 0 3.125 0 1 -0.375 850 3.125 = 272 850 3.125 = 272 (850)/(3.125)=272\frac{850}{3.125}=2728503.125=272
x 1 x 1 x_(1)x_1x1 1500 550 1 0.625 0 0 0.125 550 0.625 = 880 550 0.625 = 880 (550)/(0.625)=880\frac{550}{0.625}=8805500.625=880
Z = 8 2 5 0 0 0 Z = 8 2 5 0 0 0 Z=825000Z=\mathbf{8 2 5 0 0 0}Z=825000 Z j Z j Z_(j)Z_jZj 1500 937.5 0 0 187.5
C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj 0 262.5 ↑ 262.5 ↑ 262.5 uarr262.5 \uparrow262.5↑ 0 0 -187.5
Iteration-2 C_(j) 1500 1200 0 0 0 B C_(B) X_(B) x_(1) x_(2) s_(1) s_(2) S_(3) ” MinRatio (X_(B))/(x_(2))” s_(1) 0 300 0 (1.25) 1 0 -0.75 (300)/(1.25)=240 rarr S_(2) 0 850 0 3.125 0 1 -0.375 (850)/(3.125)=272 x_(1) 1500 550 1 0.625 0 0 0.125 (550)/(0.625)=880 Z=825000 Z_(j) 1500 937.5 0 0 187.5 C_(j)-Z_(j) 0 262.5 uarr 0 0 -187.5 | Iteration-2 | | $C_j$ | 1500 | 1200 | 0 | 0 | 0 | | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $B$ | $C_B$ | $X_B$ | $x_1$ | $x_2$ | $s_1$ | $s_2$ | $S_3$ | $\begin{array}{c}\text { MinRatio } \\ \frac{X_B}{x_2}\end{array}$ | | $s_1$ | 0 | 300 | 0 | (1.25) | 1 | 0 | -0.75 | $\frac{300}{1.25}=240 \rightarrow$ | | $S_2$ | 0 | 850 | 0 | 3.125 | 0 | 1 | -0.375 | $\frac{850}{3.125}=272$ | | $x_1$ | 1500 | 550 | 1 | 0.625 | 0 | 0 | 0.125 | $\frac{550}{0.625}=880$ | | $Z=\mathbf{8 2 5 0 0 0}$ | | $Z_j$ | 1500 | 937.5 | 0 | 0 | 187.5 | | | | | $C_j-Z_j$ | 0 | $262.5 \uparrow$ | 0 | 0 | -187.5 | |
Positive maximum C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj is 262.5 and its column index is 2 . So, the entering variable is x 2 x 2 x_(2)x_2x2.
Minimum ratio is 240 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 1.25 .
Entering = x 2 = x 2 =x_(2)=x_2=x2, Departing = S 1 = S 1 =S_(1)=S_1=S1, Key Element = 1.25 = 1.25 =1.25=1.25=1.25
R 1 ( new ) = R 1 ( old ) ÷ 1.25 R 2 ( new ) = R 2 ( old ) − 3.125 R 1 ( new ) R 3 ( new ) = R 3 ( old ) − 0.625 R 1 ( new ) R 1 (  new  ) = R 1 (  old  ) ÷ 1.25 R 2 (  new  ) = R 2 (  old  ) − 3.125 R 1 (  new  ) R 3 (  new  ) = R 3 (  old  ) − 0.625 R 1 (  new  ) {:[R_(1)(” new “)=R_(1)(” old “)-:1.25],[R_(2)(” new “)=R_(2)(” old “)-3.125R_(1)(” new “)],[R_(3)(” new “)=R_(3)(” old “)-0.625R_(1)(” new “)]:}\begin{aligned} & R_1(\text { new })=R_1(\text { old }) \div 1.25 \\ & R_2(\text { new })=R_2(\text { old })-3.125 R_1(\text { new }) \\ & R_3(\text { new })=R_3(\text { old })-0.625 R_1(\text { new }) \end{aligned}R1( new )=R1( old )÷1.25R2( new )=R2( old )−3.125R1( new )R3( new )=R3( old )−0.625R1( new )
Iteration-3 C j C j C_(j)C_jCj 1500 1200 0 0 0
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 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}}S3 MinRatio
x 2 x 2 x_(2)x_2x2 1200 240 0 1 0.8 0 -0.6
S 2 S 2 S_(2)S_2S2 0 100 0 0 -2.5 1 1.5
x 1 x 1 x_(1)x_1x1 1500 400 1 0 -0.5 0 0.5
Z = 8 8 8 0 0 0 Z = 8 8 8 0 0 0 Z=888000\boldsymbol{Z}=\mathbf{8 8 8 0 0 0}Z=888000 Z j Z j Z_(j)\boldsymbol{Z}_{\boldsymbol{j}}Zj 1 5 0 0 1 5 0 0 1500\mathbf{1 5 0 0}1500 1 2 0 0 1 2 0 0 1200\mathbf{1 2 0 0}1200 2 1 0 2 1 0 210\mathbf{2 1 0}210 0 0 0\mathbf{0}0 3 0 3 0 30\mathbf{3 0}30
C j − Z j C j − Z j C_(j)-Z_(j)C_j-Z_jCj−Zj 0 0 -210 0 -30
Iteration-3 C_(j) 1500 1200 0 0 0 B C_(B) X_(B) x_(1) x_(2) S_(1) S_(2) S_(3) MinRatio x_(2) 1200 240 0 1 0.8 0 -0.6 S_(2) 0 100 0 0 -2.5 1 1.5 x_(1) 1500 400 1 0 -0.5 0 0.5 Z=888000 Z_(j) 1500 1200 210 0 30 C_(j)-Z_(j) 0 0 -210 0 -30 | Iteration-3 | | $C_j$ | 1500 | 1200 | 0 | 0 | 0 | | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | $\boldsymbol{B}$ | $\boldsymbol{C}_{\boldsymbol{B}}$ | $\boldsymbol{X}_{\boldsymbol{B}}$ | $\boldsymbol{x}_{\mathbf{1}}$ | $\boldsymbol{x}_{\mathbf{2}}$ | $\boldsymbol{S}_{\mathbf{1}}$ | $\boldsymbol{S}_{\mathbf{2}}$ | $\boldsymbol{S}_{\mathbf{3}}$ | MinRatio | | $x_2$ | 1200 | 240 | 0 | 1 | 0.8 | 0 | -0.6 | | | $S_2$ | 0 | 100 | 0 | 0 | -2.5 | 1 | 1.5 | | | $x_1$ | 1500 | 400 | 1 | 0 | -0.5 | 0 | 0.5 | | | $\boldsymbol{Z}=\mathbf{8 8 8 0 0 0}$ | | $\boldsymbol{Z}_{\boldsymbol{j}}$ | $\mathbf{1 5 0 0}$ | $\mathbf{1 2 0 0}$ | $\mathbf{2 1 0}$ | $\mathbf{0}$ | $\mathbf{3 0}$ | | | | | $C_j-Z_j$ | 0 | 0 | -210 | 0 | -30 | |
Since all C j − Z j ≤ 0 C j − Z j ≤ 0 C_(j)-Z_(j) <= 0C_j-Z_j \leq 0Cj−Zj≤0
Hence, optimal solution is arrived with value of variables as :
x 1 = 400 , x 2 = 240 x 1 = 400 , x 2 = 240 x_(1)=400,x_(2)=240x_1=400, x_2=240x1=400,x2=240
Max ⁡ Z = 888000 Max ⁡ Z = 888000 Max Z=888000\operatorname{Max} Z=888000Max⁡Z=888000

Conclusion

Through iterative calculations and optimizations, the LP problem reached an optimal solution where:
  • x 1 = 400 x 1 = 400 x_(1)=400x_1=400x1=400 tons (mixture in the ratio 3 : 3 : 4 3 : 3 : 4 3:3:43: 3: 43:3:4 )
  • x 2 = 240 x 2 = 240 x_(2)=240x_2=240x2=240 tons (mixture in the ratio 2 : 4 : 2 2 : 4 : 2 2:4:22: 4: 22:4:2 )
    Maximum Profit:
    The maximum profit that the firm can achieve with this optimal solution is: Max ⁡ Z = 888 , 000 Max ⁡ Z = 888 , 000 Max Z=888,000\operatorname{Max} Z=888,000Max⁡Z=888,000 (in Rs.)
Verified Answer
5/5
\(cos\:2\theta =2\:cos^2\theta -1\)
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