Free BECC-105 Solved Assignment | July 2023-January 2024 | INTERMEDIATE MICROECONOMICS – I | IGNOU

BECC-105 Solved Assignment

INTERMEDIATE MICROECONOMICS – I

Assignment 1

1.(a) Explain the term Budget constraint? If the income of the consumer increases and one of the prices of the commodity decreases at the same time, will the consumer necessarily be at least as well off? Illustrate with diagram.
(b) What is utility function? Distinguish between direct utility function and indirect utility function. State the mathematical expressions of utility function of perfect substitutes and perfect compliments.
2 (a) What do you mean by ‘Cost minimisation’. Explain the various approaches of cost minimisation, Give illustration in support of your answer.
(b) Given the total cost function
TC = Q 3 5 Q 2 + 60 Q TC = Q 3 5 Q 2 + 60 Q TC=Q^(3)-5Q^(2)+60Q\mathrm{TC}=\mathrm{Q}^3-5 \mathrm{Q}^2+60 \mathrm{Q}TC=Q35Q2+60Q
Find (i) The Average cost function (ii) The critical value at which AC is minimized (iii) The Marginal Cost

Assignment 2

  1. Given the profit function
    π = 160 x 3 x 2 2 x y 2 y 2 + 120 y 18 π = 160 x 3 x 2 2 x y 2 y 2 + 120 y 18 pi=160 x-3x^(2)-2xy-2y^(2)+120 y-18\pi=160 x-3 x^2-2 x y-2 y^2+120 y-18π=160x3x22xy2y2+120y18 for a firm producing two goods x and y
    Find out (i) the maximizing profits
    (ii) test the second order condition.
  2. Distinguish between price elasticity of demand and income elasticity of demand. Given Q = 700 Q = 700 Q=700\mathrm{Q}=700Q=700
2 P + 0.02 y 2 P + 0.02 y 2P+0.02y2 \mathrm{P}+0.02 \mathrm{y}2P+0.02y Where P = 25 P = 25 P=25\mathrm{P}=25P=25, and y = 500 y = 500 y=500\mathrm{y}=500y=500
Find out (i) the price elasticity of demand and
(ii) Income elasticity of demand.
5. Do you think that Walrasian equilibrium is Pareto optimal? Give reasons and proof in support of your answer.

Assignment 3

  1. Explain the properties of preferences with example.
  2. What is consumer surplus? State the relationship between consumer surplus, compensating variation and equivalent variation.
  3. What is risk aversion? How does insurance help in reducing risk? Illustrate.
  4. What is CES production function? How does CES production function approaches a Leontief Production function?
  5. Make distinction between any three of the following:
    (i) Concave function and convex function.
    (ii) Expected value and Expected utility.
    (iii) General equilibrium and partial equilibrium.
    (iv) Marginal Rate of Substitution and Marginal Rate of Technical Substitution.

Expert Answer:

Assignment 1

Question:-01(a)

Explain the term Budget constraint? If the income of the consumer increases and one of the prices of the commodity decreases at the same time, will the consumer necessarily be at least as well off? Illustrate with diagram.

Answer:

Budget Constraint

A budget constraint represents the combination of goods and services that a consumer can purchase given their income and the prices of those goods and services. It is typically depicted as a line on a graph where one axis represents the quantity of one good and the other axis represents the quantity of another good. The budget constraint shows all the possible combinations of two goods that a consumer can buy when spending all of their income.
Mathematically, the budget constraint is written as:
P 1 X 1 + P 2 X 2 = I P 1 X 1 + P 2 X 2 = I P_(1)*X_(1)+P_(2)*X_(2)=IP_1 \cdot X_1 + P_2 \cdot X_2 = IP1X1+P2X2=I
Where:
  • P 1 P 1 P_(1)P_1P1 and P 2 P 2 P_(2)P_2P2 are the prices of goods 1 and 2.
  • X 1 X 1 X_(1)X_1X1 and X 2 X 2 X_(2)X_2X2 are the quantities of goods 1 and 2.
  • I I III is the consumer’s income.

Impact of Income Increase and Price Decrease

If the consumer’s income increases, their budget constraint shifts outward, allowing them to afford more of both goods. Simultaneously, if the price of one of the commodities decreases, the slope of the budget line will change, becoming less steep for that good. These two effects combined tend to increase the consumer’s purchasing power, allowing them to afford more goods or a better combination of goods than before.

Illustration with a Diagram

In the diagram below:
  • The initial budget constraint (BC1) represents the consumer’s original income and prices.
  • When the consumer’s income increases and the price of one good decreases, the budget constraint shifts outward (BC2).
This shift allows the consumer to move to a higher indifference curve, representing a higher level of satisfaction or utility, assuming rational consumer behavior. In most cases, the consumer will be at least as well off as before.
original image
The diagram above shows the impact of an increase in income and a decrease in the price of one good on the budget constraint.
  • The blue line represents the initial budget constraint, which shows the combinations of goods the consumer can afford with their initial income and prices.
  • The green dashed line represents the new budget constraint after the consumer’s income increases and the price of one good decreases.
As we can see, the new budget constraint (green) lies outward compared to the initial one (blue), meaning the consumer can afford more of both goods. This shift generally implies that the consumer is at least as well off, and potentially better off, since they can now access more preferred combinations of goods.




Question:-01(b)

What is utility function? Distinguish between direct utility function and indirect utility function. State the mathematical expressions of utility function of perfect substitutes and perfect complements.

Answer:

Utility Function

A utility function represents a consumer’s preferences by assigning a numerical value (utility) to different bundles of goods. It measures the satisfaction or happiness a consumer derives from consuming certain quantities of goods or services.
The utility function is usually expressed as:
U ( X 1 , X 2 , . . . , X n ) U ( X 1 , X 2 , . . . , X n ) U(X_(1),X_(2),…,X_(n))U(X_1, X_2, …, X_n)U(X1,X2,...,Xn)
Where:
  • X 1 , X 2 , . . . , X n X 1 , X 2 , . . . , X n X_(1),X_(2),…,X_(n)X_1, X_2, …, X_nX1,X2,...,Xn represent the quantities of different goods consumed.
  • U U UUU is the utility derived from the consumption bundle ( X 1 , X 2 , . . . , X n ) ( X 1 , X 2 , . . . , X n ) (X_(1),X_(2),…,X_(n))(X_1, X_2, …, X_n)(X1,X2,...,Xn).

Direct Utility Function vs. Indirect Utility Function

  1. Direct Utility Function:
    • The direct utility function describes the utility derived directly from the consumption of goods and services. It focuses on how much utility the consumer gets from different consumption bundles, assuming they have certain quantities of goods.
    • It can be expressed as: U ( X 1 , X 2 , . . . , X n ) U ( X 1 , X 2 , . . . , X n ) U(X_(1),X_(2),…,X_(n))U(X_1, X_2, …, X_n)U(X1,X2,...,Xn)
    • The direct utility function does not consider the prices of goods or the consumer’s income but purely reflects their preferences.
  2. Indirect Utility Function:
    • The indirect utility function describes the maximum utility that a consumer can achieve given their income and the prices of goods. It takes into account the consumer’s budget constraint.
    • The indirect utility function is derived from the optimization of the consumer’s utility under the budget constraint and is expressed as: V ( I , P 1 , P 2 , . . . , P n ) V ( I , P 1 , P 2 , . . . , P n ) V(I,P_(1),P_(2),…,P_(n))V(I, P_1, P_2, …, P_n)V(I,P1,P2,...,Pn)
    • Where I I III is the consumer’s income and P 1 , P 2 , . . . , P n P 1 , P 2 , . . . , P n P_(1),P_(2),…,P_(n)P_1, P_2, …, P_nP1,P2,...,Pn are the prices of the goods.
    • It represents the highest level of utility that can be attained given income and prices.

Utility Functions for Perfect Substitutes and Perfect Complements

1. Perfect Substitutes:

  • Goods are considered perfect substitutes when a consumer is willing to substitute one good for another at a constant rate.
  • The utility function for perfect substitutes is usually of the form: U ( X 1 , X 2 ) = a X 1 + b X 2 U ( X 1 , X 2 ) = a X 1 + b X 2 U(X_(1),X_(2))=aX_(1)+bX_(2)U(X_1, X_2) = aX_1 + bX_2U(X1,X2)=aX1+bX2
  • Where a a aaa and b b bbb are constants representing the rate at which one good can be substituted for the other without changing the overall utility.

2. Perfect Complements:

  • Goods are considered perfect complements when they are always consumed together in fixed proportions, such as left and right shoes.
  • The utility function for perfect complements is typically of the form: U ( X 1 , X 2 ) = min ( a X 1 , b X 2 ) U ( X 1 , X 2 ) = min ( a X 1 , b X 2 ) U(X_(1),X_(2))=min(aX_(1),bX_(2))U(X_1, X_2) = \min(aX_1, bX_2)U(X1,X2)=min(aX1,bX2)
  • Here, a a aaa and b b bbb are constants that indicate the fixed proportion in which the goods must be consumed to generate utility.

Summary of Mathematical Expressions

  1. Perfect Substitutes Utility Function: U ( X 1 , X 2 ) = a X 1 + b X 2 U ( X 1 , X 2 ) = a X 1 + b X 2 U(X_(1),X_(2))=aX_(1)+bX_(2)U(X_1, X_2) = aX_1 + bX_2U(X1,X2)=aX1+bX2
  2. Perfect Complements Utility Function: U ( X 1 , X 2 ) = min ( a X 1 , b X 2 ) U ( X 1 , X 2 ) = min ( a X 1 , b X 2 ) U(X_(1),X_(2))=min(aX_(1),bX_(2))U(X_1, X_2) = \min(aX_1, bX_2)U(X1,X2)=min(aX1,bX2)
These functions represent different consumer preferences regarding how they value goods either as substitutes or complements.




Question:-02(a)

What do you mean by ‘Cost minimisation’? Explain the various approaches of cost minimisation, Give illustration in support of your answer.

Answer:

Cost Minimization

Cost minimization refers to the process by which a firm determines the most cost-efficient combination of inputs (such as labor, capital, raw materials, etc.) to produce a given level of output. The goal is to minimize the total cost of production while maintaining the desired output level.
Firms are assumed to be profit-maximizing entities, and minimizing production costs is a critical step toward maximizing profits. This concept is central in economics and managerial decision-making, where firms seek to produce at the lowest possible cost for a given level of output.

Approaches to Cost Minimization

There are two main approaches to cost minimization:
  1. Isoquant-Isocost Approach
  2. Marginal Product Approach

1. Isoquant-Isocost Approach

This is a graphical approach to cost minimization. It involves using two key concepts: isoquants and isocost lines.
  • Isoquant: An isoquant represents all combinations of two inputs (say labor and capital) that can produce a given level of output. It is similar to an indifference curve in utility analysis but applies to production.
  • Isocost Line: An isocost line represents all combinations of two inputs that have the same total cost. The equation for an isocost line is:
    C = w L + r K C = w L + r K C=wL+rKC = wL + rKC=wL+rK
    Where:
    • C C CCC is the total cost,
    • w w www is the wage rate (cost of labor),
    • r r rrr is the rental rate of capital (cost of capital),
    • L L LLL is the quantity of labor,
    • K K KKK is the quantity of capital.
The goal of the firm is to find the point where an isoquant (representing a given output level) is tangent to an isocost line (representing the minimum cost of inputs). This tangency condition is where the firm is producing the given output at the minimum possible cost.
The mathematical condition for cost minimization is:
M P L w = M P K r M P L w = M P K r (MP_(L))/(w)=(MP_(K))/(r)\frac{MP_L}{w} = \frac{MP_K}{r}MPLw=MPKr
Where:
  • M P L M P L MP_(L)MP_LMPL is the marginal product of labor,
  • M P K M P K MP_(K)MP_KMPK is the marginal product of capital.
This condition implies that the firm should allocate its inputs such that the marginal product per dollar spent on each input is equal.

2. Marginal Product Approach

This approach is based on comparing the marginal products of the inputs relative to their costs. The firm will adjust its input usage until the marginal product per dollar spent is equalized across all inputs. This method is often applied analytically rather than graphically and is based on the following rules:
  • If the marginal product per dollar of labor is higher than that of capital, the firm should hire more labor and use less capital.
  • If the marginal product per dollar of capital is higher than that of labor, the firm should use more capital and less labor.
The goal is to balance the cost per unit of output across inputs.

Illustration of Cost Minimization

Consider a firm that produces a fixed level of output (say 100 units) using labor and capital. The firm’s objective is to minimize the cost of producing this output level.
  • Labor costs: $10 per unit.
  • Capital costs: $20 per unit.
  • The firm needs to decide how much labor and capital to use to produce 100 units of output.
Step 1: Isoquant and Isocost Curves
  • The isoquant curve shows different combinations of labor and capital that can produce 100 units of output.
  • The isocost line shows different combinations of labor and capital that can be hired for a given total cost.
Step 2: Tangency Condition
  • The cost-minimizing point occurs where the isoquant for 100 units of output is tangent to the isocost line. At this point, the marginal rate of technical substitution (MRTS) of labor for capital is equal to the ratio of the prices of labor and capital: M P L M P K = w r M P L M P K = w r (MP_(L))/(MP_(K))=(w)/(r)\frac{MP_L}{MP_K} = \frac{w}{r}MPLMPK=wr
Step 3: Optimal Input Mix
  • Suppose at the tangency point, the firm uses 5 units of labor and 3 units of capital to produce the 100 units of output. The total cost at this point is: Total Cost = w L + r K = ( 10 × 5 ) + ( 20 × 3 ) = 50 + 60 = 110 Total Cost = w L + r K = ( 10 × 5 ) + ( 20 × 3 ) = 50 + 60 = 110 “Total Cost”=wL+rK=(10 xx5)+(20 xx3)=50+60=110\text{Total Cost} = wL + rK = (10 \times 5) + (20 \times 3) = 50 + 60 = 110Total Cost=wL+rK=(10×5)+(20×3)=50+60=110
  • This combination minimizes the total cost of production for 100 units of output.
original image
The diagram above illustrates the isoquant and isocost lines in the cost minimization problem:
  • The blue curve is the isoquant, representing all the combinations of labor and capital that produce a given level of output (100 units in this case).
  • The green dashed line is the isocost line, which shows all combinations of labor and capital that result in a total cost of $110.
The point where the isoquant curve is tangent to the isocost line represents the optimal combination of labor and capital that minimizes cost for the given level of output. This is where the firm achieves cost minimization while still producing the desired output.
By balancing the costs of inputs according to their marginal productivity, the firm ensures it is operating efficiently.




Question:-02(b)

Given the total cost function
TC = Q 3 5 Q 2 + 60 Q TC = Q 3 5 Q 2 + 60 Q TC=Q^(3)-5Q^(2)+60Q\mathrm{TC}=\mathrm{Q}^3-5 \mathrm{Q}^2+60 \mathrm{Q}TC=Q35Q2+60Q
Find:
(i) The Average cost function
(ii) The critical value at which AC is minimized
(iii) The Marginal Cost

Answer:

Let’s solve the given problem step by step:

Given Total Cost Function:

T C = Q 3 5 Q 2 + 60 Q T C = Q 3 5 Q 2 + 60 Q TC=Q^(3)-5Q^(2)+60 QTC = Q^3 – 5Q^2 + 60QTC=Q35Q2+60Q

(i) The Average Cost Function

The Average Cost (AC) function is calculated by dividing the total cost (TC) by the quantity (Q):
A C = T C Q A C = T C Q AC=(TC)/(Q)AC = \frac{TC}{Q}AC=TCQ
Substituting the expression for T C T C TCTCTC:
A C = Q 3 5 Q 2 + 60 Q Q A C = Q 3 5 Q 2 + 60 Q Q AC=(Q^(3)-5Q^(2)+60 Q)/(Q)AC = \frac{Q^3 – 5Q^2 + 60Q}{Q}AC=Q35Q2+60QQ
Simplifying the equation:
A C = Q 2 5 Q + 60 A C = Q 2 5 Q + 60 AC=Q^(2)-5Q+60AC = Q^2 – 5Q + 60AC=Q25Q+60
Thus, the Average Cost function is:
A C = Q 2 5 Q + 60 A C = Q 2 5 Q + 60 AC=Q^(2)-5Q+60AC = Q^2 – 5Q + 60AC=Q25Q+60

(ii) The Critical Value at Which AC is Minimized

To find the critical value, we need to minimize the average cost function. To do this, we differentiate the AC function with respect to Q Q QQQ and set the derivative equal to zero.
  1. Differentiate the AC function:
d ( A C ) d Q = 2 Q 5 d ( A C ) d Q = 2 Q 5 (d(AC))/(dQ)=2Q-5\frac{d(AC)}{dQ} = 2Q – 5d(AC)dQ=2Q5
  1. Set the derivative equal to zero to find the critical point:
2 Q 5 = 0 2 Q 5 = 0 2Q-5=02Q – 5 = 02Q5=0
Solving for Q Q QQQ:
Q = 5 2 = 2.5 Q = 5 2 = 2.5 Q=(5)/(2)=2.5Q = \frac{5}{2} = 2.5Q=52=2.5
Thus, the critical value at which AC is minimized is Q = 2.5 Q = 2.5 Q=2.5Q = 2.5Q=2.5.

(iii) The Marginal Cost

The Marginal Cost (MC) function is the derivative of the total cost (TC) function with respect to Q Q QQQ:
M C = d ( T C ) d Q M C = d ( T C ) d Q MC=(d(TC))/(dQ)MC = \frac{d(TC)}{dQ}MC=d(TC)dQ
Differentiating the total cost function:
M C = d d Q ( Q 3 5 Q 2 + 60 Q ) = 3 Q 2 10 Q + 60 M C = d d Q ( Q 3 5 Q 2 + 60 Q ) = 3 Q 2 10 Q + 60 MC=(d)/(dQ)(Q^(3)-5Q^(2)+60 Q)=3Q^(2)-10 Q+60MC = \frac{d}{dQ}(Q^3 – 5Q^2 + 60Q) = 3Q^2 – 10Q + 60MC=ddQ(Q35Q2+60Q)=3Q210Q+60
Thus, the Marginal Cost function is:
M C = 3 Q 2 10 Q + 60 M C = 3 Q 2 10 Q + 60 MC=3Q^(2)-10 Q+60MC = 3Q^2 – 10Q + 60MC=3Q210Q+60

Summary of Results:

  1. Average Cost function (AC): A C = Q 2 5 Q + 60 A C = Q 2 5 Q + 60 AC=Q^(2)-5Q+60AC = Q^2 – 5Q + 60AC=Q25Q+60
  2. Critical value at which AC is minimized: Q = 2.5 Q = 2.5 Q=2.5Q = 2.5Q=2.5
  3. Marginal Cost function (MC): M C = 3 Q 2 10 Q + 60 M C = 3 Q 2 10 Q + 60 MC=3Q^(2)-10 Q+60MC = 3Q^2 – 10Q + 60MC=3Q210Q+60




Assignment 2

Question:-03

Given the profit function
π = 160 x 3 x 2 2 x y 2 y 2 + 120 y 18 π = 160 x 3 x 2 2 x y 2 y 2 + 120 y 18 pi=160 x-3x^(2)-2xy-2y^(2)+120 y-18\pi=160 x-3 x^2-2 x y-2 y^2+120 y-18π=160x3x22xy2y2+120y18 for a firm producing two goods x and y
Find out:
(i) the maximizing profits
(ii) test the second order condition.

Answer:

To solve this problem, we need to maximize the profit function given by:
π = 160 x 3 x 2 2 x y 2 y 2 + 120 y 18 π = 160 x 3 x 2 2 x y 2 y 2 + 120 y 18 pi=160 x-3x^(2)-2xy-2y^(2)+120 y-18\pi = 160x – 3x^2 – 2xy – 2y^2 + 120y – 18π=160x3x22xy2y2+120y18

(i) Finding the Maximum Profit

To find the maximizing profits, we first calculate the first-order conditions by taking the partial derivatives of the profit function with respect to x x xxx and y y yyy and setting them equal to zero.
  1. Partial derivative with respect to x x xxx:
π x = 160 6 x 2 y = 0 π x = 160 6 x 2 y = 0 (del pi)/(del x)=160-6x-2y=0\frac{\partial \pi}{\partial x} = 160 – 6x – 2y = 0πx=1606x2y=0
160 6 x 2 y = 0 (Equation 1) 160 6 x 2 y = 0 (Equation 1) 160-6x-2y=0quad(Equation 1)160 – 6x – 2y = 0 \quad \text{(Equation 1)}1606x2y=0(Equation 1)
  1. Partial derivative with respect to y y yyy:
π y = 120 2 x 4 y = 0 π y = 120 2 x 4 y = 0 (del pi)/(del y)=120-2x-4y=0\frac{\partial \pi}{\partial y} = 120 – 2x – 4y = 0πy=1202x4y=0
120 2 x 4 y = 0 (Equation 2) 120 2 x 4 y = 0 (Equation 2) 120-2x-4y=0quad(Equation 2)120 – 2x – 4y = 0 \quad \text{(Equation 2)}1202x4y=0(Equation 2)
Now, solve these two equations simultaneously to find the values of x x xxx and y y yyy.
Step 1: Solve Equation 2 for x x xxx:
x = 60 2 y x = 60 2 y x=60-2yx = 60 – 2yx=602y
Step 2: Substitute this into Equation 1:
160 6 ( 60 2 y ) 2 y = 0 160 6 ( 60 2 y ) 2 y = 0 160-6(60-2y)-2y=0160 – 6(60 – 2y) – 2y = 01606(602y)2y=0
160 360 + 12 y 2 y = 0 160 360 + 12 y 2 y = 0 160-360+12 y-2y=0160 – 360 + 12y – 2y = 0160360+12y2y=0
200 + 10 y = 0 200 + 10 y = 0 -200+10 y=0-200 + 10y = 0200+10y=0
y = 20 y = 20 y=20y = 20y=20
Step 3: Substitute y = 20 y = 20 y=20y = 20y=20 back into the expression for x x xxx:
x = 60 2 ( 20 ) = 20 x = 60 2 ( 20 ) = 20 x=60-2(20)=20x = 60 – 2(20) = 20x=602(20)=20
Thus, the maximizing values of x x xxx and y y yyy are x = 20 x = 20 x=20x = 20x=20 and y = 20 y = 20 y=20y = 20y=20.
Now, substitute these values into the profit function to find the maximum profit:
π = 160 ( 20 ) 3 ( 20 ) 2 2 ( 20 ) ( 20 ) 2 ( 20 ) 2 + 120 ( 20 ) 18 π = 160 ( 20 ) 3 ( 20 ) 2 2 ( 20 ) ( 20 ) 2 ( 20 ) 2 + 120 ( 20 ) 18 pi=160(20)-3(20)^(2)-2(20)(20)-2(20)^(2)+120(20)-18\pi = 160(20) – 3(20)^2 – 2(20)(20) – 2(20)^2 + 120(20) – 18π=160(20)3(20)22(20)(20)2(20)2+120(20)18
π = 3200 1200 800 800 + 2400 18 π = 3200 1200 800 800 + 2400 18 pi=3200-1200-800-800+2400-18\pi = 3200 – 1200 – 800 – 800 + 2400 – 18π=32001200800800+240018
π = 4782 π = 4782 pi=4782\pi = 4782π=4782
Thus, the maximum profit is π = 4782 π = 4782 pi=4782\pi = 4782π=4782.

(ii) Testing the Second-Order Condition

To test the second-order condition, we need to verify that the profit function is concave at the critical points x = 20 x = 20 x=20x = 20x=20 and y = 20 y = 20 y=20y = 20y=20. This involves checking the Hessian matrix of second-order partial derivatives.
  1. Second-order partial derivatives:
2 π x 2 = 6 2 π x 2 = 6 (del^(2)pi)/(delx^(2))=-6\frac{\partial^2 \pi}{\partial x^2} = -62πx2=6
2 π y 2 = 4 2 π y 2 = 4 (del^(2)pi)/(dely^(2))=-4\frac{\partial^2 \pi}{\partial y^2} = -42πy2=4
2 π x y = 2 π y x = 2 2 π x y = 2 π y x = 2 (del^(2)pi)/(del x del y)=(del^(2)pi)/(del y del x)=-2\frac{\partial^2 \pi}{\partial x \partial y} = \frac{\partial^2 \pi}{\partial y \partial x} = -22πxy=2πyx=2
The Hessian matrix H H HHH is:
H = ( 6 2 2 4 ) H = 6 2 2 4 H=([-6,-2],[-2,-4])H = \begin{pmatrix} -6 & -2 \\ -2 & -4 \end{pmatrix}H=(6224)
  1. Determinant of the Hessian:
Det ( H ) = ( 6 ) ( 4 ) ( 2 ) ( 2 ) = 24 4 = 20 Det ( H ) = ( 6 ) ( 4 ) ( 2 ) ( 2 ) = 24 4 = 20 “Det”(H)=(-6)(-4)-(-2)(-2)=24-4=20\text{Det}(H) = (-6)(-4) – (-2)(-2) = 24 – 4 = 20Det(H)=(6)(4)(2)(2)=244=20
  1. Principal minors:
  • The first principal minor is H 11 = 6 H 11 = 6 H_(11)=-6H_{11} = -6H11=6, which is negative.
  • The determinant of the Hessian Det ( H ) = 20 Det ( H ) = 20 “Det”(H)=20\text{Det}(H) = 20Det(H)=20, which is positive.
Since the first principal minor is negative and the determinant of the Hessian is positive, the Hessian is negative definite. This satisfies the second-order condition for a local maximum.

Conclusion

  1. The maximizing profit is π = 4782 π = 4782 pi=4782\pi = 4782π=4782 at x = 20 x = 20 x=20x = 20x=20 and y = 20 y = 20 y=20y = 20y=20.
  2. The second-order condition is satisfied, confirming that this is a maximum point.




Question:-04

Distinguish between price elasticity of demand and income elasticity of demand. Given Q = 700 Q = 700 Q=700\mathrm{Q}=700Q=700
2 P + 0.02 y 2 P + 0.02 y 2P+0.02y2 \mathrm{P}+0.02 \mathrm{y}2P+0.02y, where P = 25 P = 25 P=25\mathrm{P}=25P=25, and y = 500 y = 500 y=500\mathrm{y}=500y=500
Find out:
(i) the price elasticity of demand
(ii) Income elasticity of demand.

Answer:

Price Elasticity of Demand vs. Income Elasticity of Demand

Price Elasticity of Demand (PED) measures the responsiveness of the quantity demanded of a good to a change in its price. It is calculated as the percentage change in quantity demanded divided by the percentage change in price:
Price Elasticity of Demand (PED) = % Δ Q % Δ P = Q P P Q Price Elasticity of Demand (PED) = % Δ Q % Δ P = Q P P Q “Price Elasticity of Demand (PED)”=(%Delta Q)/(%Delta P)=((del Q)/(del P)*P)/(Q)\text{Price Elasticity of Demand (PED)} = \frac{\%\Delta Q}{\%\Delta P} = \frac{\frac{\partial Q}{\partial P} \cdot P}{Q}Price Elasticity of Demand (PED)=%ΔQ%ΔP=QPPQ
Income Elasticity of Demand (YED) measures the responsiveness of the quantity demanded of a good to a change in consumers’ income. It is calculated as the percentage change in quantity demanded divided by the percentage change in income:
Income Elasticity of Demand (YED) = % Δ Q % Δ Y = Q y y Q Income Elasticity of Demand (YED) = % Δ Q % Δ Y = Q y y Q “Income Elasticity of Demand (YED)”=(%Delta Q)/(%Delta Y)=((del Q)/(del y)*y)/(Q)\text{Income Elasticity of Demand (YED)} = \frac{\%\Delta Q}{\%\Delta Y} = \frac{\frac{\partial Q}{\partial y} \cdot y}{Q}Income Elasticity of Demand (YED)=%ΔQ%ΔY=QyyQ

Given Information:

Q = 700 2 P + 0.02 y Q = 700 2 P + 0.02 y Q=700-2P+0.02 yQ = 700 – 2P + 0.02yQ=7002P+0.02y
Where:
  • P = 25 P = 25 P=25P = 25P=25
  • y = 500 y = 500 y=500y = 500y=500
Let’s first compute the values of Q Q QQQ, Q P Q P (del Q)/(del P)\frac{\partial Q}{\partial P}QP, and Q y Q y (del Q)/(del y)\frac{\partial Q}{\partial y}Qy using the given information.

Step 1: Calculate the Quantity Demanded (Q)

Substituting P = 25 P = 25 P=25P = 25P=25 and y = 500 y = 500 y=500y = 500y=500 into the demand function:
Q = 700 2 ( 25 ) + 0.02 ( 500 ) Q = 700 2 ( 25 ) + 0.02 ( 500 ) Q=700-2(25)+0.02(500)Q = 700 – 2(25) + 0.02(500)Q=7002(25)+0.02(500)
Q = 700 50 + 10 = 660 Q = 700 50 + 10 = 660 Q=700-50+10=660Q = 700 – 50 + 10 = 660Q=70050+10=660

(i) Price Elasticity of Demand (PED)

  1. First, calculate the partial derivative of Q Q QQQ with respect to P P PPP:
Q P = 2 Q P = 2 (del Q)/(del P)=-2\frac{\partial Q}{\partial P} = -2QP=2
  1. Now, use the formula for Price Elasticity of Demand:
PED = Q P P Q PED = Q P P Q “PED”=((del Q)/(del P)*P)/(Q)\text{PED} = \frac{\frac{\partial Q}{\partial P} \cdot P}{Q}PED=QPPQ
Substituting the values:
PED = 2 25 660 = 50 660 0.0758 PED = 2 25 660 = 50 660 0.0758 “PED”=(-2*25)/(660)=(-50)/(660)~~-0.0758\text{PED} = \frac{-2 \cdot 25}{660} = \frac{-50}{660} \approx -0.0758PED=225660=506600.0758
Thus, the price elasticity of demand is approximately 0.076 0.076 -0.076-0.0760.076.

(ii) Income Elasticity of Demand (YED)

  1. First, calculate the partial derivative of Q Q QQQ with respect to y y yyy:
Q y = 0.02 Q y = 0.02 (del Q)/(del y)=0.02\frac{\partial Q}{\partial y} = 0.02Qy=0.02
  1. Now, use the formula for Income Elasticity of Demand:
YED = Q y y Q YED = Q y y Q “YED”=((del Q)/(del y)*y)/(Q)\text{YED} = \frac{\frac{\partial Q}{\partial y} \cdot y}{Q}YED=QyyQ
Substituting the values:
YED = 0.02 500 660 = 10 660 0.0152 YED = 0.02 500 660 = 10 660 0.0152 “YED”=(0.02*500)/(660)=(10)/(660)~~0.0152\text{YED} = \frac{0.02 \cdot 500}{660} = \frac{10}{660} \approx 0.0152