ECO406 ASSIGNMENT NO. 2 SPRING 2023 || 100% RIGHT SOLUTION || MATHEMATICAL ECONOMICS || BY VuTech

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QUESTION-1

Constraints limit the options available to the decision makers. For example, consumer has to face constraints like he must consume goods within his given income. There fore, we need to find a maximum of his utility function keeping in view his given income and the prices of the goods. Utility maximization refers to the concept that individuals and firms seek to get the highest satisfaction from their economic decisions. Keeping in view the concept of budget constraint,

Find the maximum of utility function

U= Q1Q2 (Objective Function)

in the presence of the following budget constraints. 

B=P1Q1+P2Q2 (Subject to)

Where the income B= 440, P1= 20 and P2 = 4

ANSWER:

To find the maximum of the utility function U = Q1 * Q2, subject to the budget constraint B 

= P1 * Q1 + P2 * Q2, 

we need to use the method of Lagrange multipliers. This method allows us to optimize a function subject to one or more constraints.

Let's set up the Lagrangian function:

L(Q1, Q2, λ) = U - λ * (B - P1 * Q1 - P2 * Q2)

where λ is the Lagrange multiplier.

Now, we'll take the partial derivatives of L with respect to Q1, Q2, and λ and set them equal to zero to find the critical points:

∂L/∂Q1 = 0

∂L/∂Q2 = 0

∂L/∂λ = 0

Taking the partial derivative with respect to Q1:

∂L/∂Q1 = ∂(Q1 * Q2 - λ * (B - P1 * Q1 - P2 * Q2))/∂Q1

= Q2 - λ * (-P1) = 0

Taking the partial derivative with respect to Q2:

∂L/∂Q2 = ∂(Q1 * Q2 - λ * (B - P1 * Q1 - P2 * Q2))/∂Q2

= Q1 - λ * (-P2) = 0

Taking the partial derivative with respect to λ:

∂L/∂λ = B - P1 * Q1 - P2 * Q2 = 0

Now, we have a system of three equations:

Q2 + λ * P1 = 0

Q1 + λ * P2 = 0

B - P1 * Q1 - P2 * Q2 = 0

Given the values of B = 440, P1 = 20, and P2 = 4, we can solve these equations to find the values of Q1, Q2, and λ.

From equations 1 and 2:

Q2 = -λ * P1

Q1 = -λ * P2

Substitute these values into equation 3:

440 - 20 * (-λ * P2) - 4 * (-λ * P1) = 0

440 + 80λ = 0

λ = -440/80

λ = -5.5

Now, find Q1 and Q2 using the values of λ:

Q1 = -(-5.5) * 4 = 22

Q2 = -(-5.5) * 20 = 110

So, the maximum utility (U) is achieved when Q1 = 22 and Q2 = 110, while satisfying the budget constraint B = 440, P1 = 20, and P2 = 4.

QUESTION-3

Find out the partial derivatives of given profit function with respect to X and Y.

π = 70X – 5X2 – 3XY – 6Y2 + 100Y

ANSWER:

To find the partial derivatives of the given profit function π with respect to X and Y, we take the derivative of the function with respect to each variable while treating the other variable as a constant.

Partial derivative with respect to X (π_X):

To find ∂π/∂X, we differentiate the profit function with respect to X while treating Y as a constant:

π_X = d/dX (70X – 5X^2 – 3XY – 6Y^2 + 100Y)

Differentiating each term:

π_X = d/dX (70X) - d/dX (5X^2) - d/dX (3XY) - d/dX (6Y^2) + d/dX (100Y)

π_X = 70 - 10X - 3Y

Partial derivative with respect to Y (π_Y):

To find ∂π/∂Y, we differentiate the profit function with respect to Y while treating X as a constant:

π_Y = d/dY (70X – 5X^2 – 3XY – 6Y^2 + 100Y)

Differentiating each term:

π_Y = d/dY (3XY) - d/dY (6Y^2) + d/dY (100Y)

π_Y = 3X - 12Y + 100

So, the partial derivative of the profit function π with respect to X is 70 - 10X - 3Y, and the partial derivative with respect to Y is 3X - 12Y + 100.

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