# Chapter 4 Henderson and Quandt Microeconomics, Questions 4.1-4.3_ Solutions

**4-1 Construct the average and marginal product functions for, which correspond to the**

**production function q = x₂-0.2-0.8. Let x₂ = 10. At what respective values of x, will the**

**AP and MP of** **equal zero?**

**Ans. **To construct the average and marginal product functions for X_1, we need to use the following formulas:

Average Product of X_1 = q / x_1

Marginal Product of X_1 = ∂q / ∂x_1

Given the production function q = x_1x_2 - 0.2x_1^2 - 0.8x_2^2 and x_2 = 10, we can substitute x_2 = 10 and simplify to get:

q = 10x_1 - 0.2x_1^2 - 800

The average product of X_1 is:

AP_x1 = q / x_1 = (10x_1 - 0.2x_1^2 - 800) / x_1 = 10 - 0.2x_1 - 800/x_1

The marginal product of X_1 is:

MP_x1 = ∂q / ∂x_1 = 10 - 0.4x_1

To find the values of x_1 at which the AP and MP of X_1 equal zero, we need to set AP_x1 and MP_x1 equal to zero and solve for x_1:

AP_x1 = 0 => 10 - 0.2x_1 - 800/x_1 = 0

Solving for x_1, we get:

x_1 = 40

MP_x1 = 0 => 10 - 0.4x_1 = 0

Solving for x_1, we get:

x_1 = 25

Therefore, the respective values of x_1 at which the AP and MP of X_1 equal zero are x_1 = 40 and x_1 = 25.

**4-2 Determine the domain over which the production function q=100(x₁+x₂)+2012.5( is increasing and strictly concave.**

To determine the domain over which the production function q=100(x₁+x₂)+20x_1 x_2-12.5(x_1^2+x_2^2) is increasing and strictly concave, we need to check its first and second order conditions.

The first order conditions for this production function are:

∂q/∂x₁ = 100 + 20x₂ - 25x₁ = 0

∂q/∂x₂ = 100 + 20x₁ - 25x₂ = 0

Solving for x₁ and x₂, we get:

x₁ = 4 + 0.8x₂

x₂ = 4 + 0.8x₁

Substituting these expressions for x₁ and x₂ back into the production function, we get:

q = 100(8 + 1.6x₁) - 12.5(16 + 1.6x₁)^2

Simplifying, we get:

q = -12.5(1.6x₁ - 8)^2 + 800

From this equation, we can see that the production function is increasing and strictly concave over the range of x₁ and x₂ that satisfies the following conditions:

1.6x₁ - 8 ≥ 0 (because otherwise, the square term in the production function will be negative and q will be decreasing)

The Hessian matrix of the production function is negative definite, which means that its second-order partial derivatives are negative.

The Hessian matrix of the production function is:

H = [∂²q/∂x₁² ∂²q/∂x₁∂x₂]

[∂²q/∂x₂∂x₁ ∂²q/∂x₂²]

Taking the second-order partial derivatives of the production function and evaluating them at x₁ = 4 + 0.8x₂ and x₂ = 4 + 0.8x₁, we get:

∂²q/∂x₁² = -50

∂²q/∂x₁∂x₂ = 20

∂²q/∂x₂∂x₁ = 20

∂²q/∂x₂² = -50

The determinant of the Hessian matrix is:

|H| = (∂²q/∂x₁²)(∂²q/∂x₂²) - (∂²q/∂x₁∂x₂)(∂²q/∂x₂∂x₁) = (-50)(-50) - (20)(20) = 1600

Since |H| > 0 and ∂²q/∂x₁² < 0, we can conclude that the Hessian matrix is negative definite and the production function is strictly concave.

Using the constraint 1.6x₁ - 8 ≥ 0, we can solve for the domain over which the production function is increasing and strictly concave:

1.6x₁ - 8 ≥ 0

x₁ ≥ 5

Therefore, the domain over which the production function q=100(x₁+x₂)+20x_1 x_2-12.5(x_1^2+x_2^2) is increasing and strictly concave is:

{x₁ ≥ 5, x₂ = 4 + 0.8x₁} or equivalently {x₁ ≥ 5, x₂ ≥ 8 + 0.8x₁}

**4-3 Derive an input expansion path for the production function q= A(+1)^a(x₂+1)^β where**

**α, β>0.**

To derive the input expansion path for the production function q= A(x_1+1)^a(x₂+1)^β where α, β > 0, we need to find the relationship between the input prices and the optimal input quantities for a given level of output.

First, we write the cost function for the production function as:

C = w₁x₁ + w₂x₂

where w₁ and w₂ are the prices of inputs x₁ and x₂, respectively.

Then, we use the production function to express output q as a function of the input prices and quantities:

q = A(x₁+1)^a(x₂+1)^β

Taking natural logarithms of both sides, we get:

ln(q) = ln(A) + a ln(x₁+1) + β ln(x₂+1)

To find the optimal input quantities that minimize cost for a given level of output q, we maximize the profit function π = pq - C, where p is the price of output. Substituting the expression for q into the profit function, we get:

π = Ap(x₁+1)^a(x₂+1)^β - w₁x₁ - w₂x₂

Taking the first order conditions with respect to x₁ and x₂, we get:

∂π/∂x₁ = Apa(x₁+1)^(a-1)(x₂+1)^β - w₁ = 0

∂π/∂x₂ = Aβ(x₁+1)^a(x₂+1)^(β-1) - w₂ = 0

Solving for x₁ and x₂, we get:

x₁ = [(w₁/Aa)(x₂+1)^(-β)]^(1/(a-1)) - 1

x₂ = [(w₂/Aβ)(x₁+1)^(-a)]^(1/(β-1)) - 1

Substituting these expressions for x₁ and x₂ back into the production function, we get the input expansion path:

q = A[((w₁/Aa)(x₂+1)^(-β))^(a/(a-1)) + 1]^a[((w₂/Aβ)(x₁+1)^(-a))^(β/(β-1)) + 1]^β

This equation shows the optimal input quantities for producing different levels of output as a function of input prices. The input expansion path traces out the combinations of x₁ and x₂ that are optimal for producing increasing levels of output, holding input prices constant.

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