Question

4. Suppose that f(x, X) = (Xia + X2"),a, what is the associated cost function?

Answer 1

Total cost TC = w₁x₁ + w₂x₂

Cost function can be derived by solving the following optimization problem -

Min $w_1x_1 + w_2x_2$

such that $f(x_1, x_2 ) = (x_1^a + x_2^a)^{1/a}$

This can be solved using Lagrange Multiplier method.

1) Setting up Lagrangian -

$L = w_1x_1 + w_2x_2 + \lambda(Q - (x_1^a + x_2^a)^{1/a})$

2) Using first order conditions -

$\frac{\partial }{\partial x_1}L = 0 \Rightarrow w_1- \frac{\lambda}{a}(x_1^a + x_2^a)^{(1/a)-1}*ax_1^{a-1} = 0$ (Differentiating using chain rule)

${\lambda}(x_1^a + x_2^a)^{(1/a)-1}*x_1^{a-1} = w_1$ (1)

$\frac{\partial }{\partial x_2}L = 0 \Rightarrow w_2- \frac{\lambda}{a}(x_1^a + x_2^a)^{(1/a)-1}*ax_2^{a-1} = 0$

${\lambda}(x_1^a + x_2^a)^{(1/a)-1}*x_2^{a-1} = w_2$ (2)

$\frac{\partial }{\partial \lambda}L = 0 \Rightarrow Q= (x_1^a + x_2^a)^{(1/a)}$ (3)

3) Solving the equations -

Dividing (1) by (2), we get:

$(\frac{x_1}{x_2})^{a-1} = \frac{w_1}{w_2}$   

$\Rightarrow (\frac{x_1}{x_2}) = (\frac{w_1}{w_2})^{\frac{1}{a-1}}$$\Rightarrow x_1^a = (\frac{w_1}{w_2})^{\frac{a}{a-1}}x_2^a$ (4)

Substituting this in (3), we get:

$Q = [(\frac{w_1}{w_2})^{\frac{a}{a-1}}x_2^a +x_2^a ]^{1/a} \Rightarrow Q = x_2[1 + (\frac{w_1}{w_2})^{\frac{a}{a-1}}] \Rightarrow x_2 = \frac{Q}{[1 + (\frac{w_1}{w_2})^{\frac{a}{a-1}}]}$

Also $x_1 = x_2(\frac{w_1}{w_2})^{\frac{1}{a-1}}$

Thus $x_1 = \frac{Q}{[1 + (\frac{w_1}{w_2})^{\frac{a}{a-1}}]}(\frac{w_1}{w_2})^{\frac{1}{a-1}}$

Thus, cost function is -

$C(w_1,w_2, Q) = w_1x_1(w_1,w_2, Q) + w_2x_2(w_2,w_2, Q)$

$= w_1 \frac{Q}{[1 + (\frac{w_1}{w_2})^{\frac{a}{a-1}}]}(\frac{w_1}{w_2})^{\frac{1}{a-1}} + w_2\frac{Q}{[1 + (\frac{w_1}{w_2})^{\frac{a}{a-1}}]}$