-2
$\begingroup$

I’ve been given this true false question: Consider the minimization of wL + rK given F(K, L) $\geq$ Q with F(K, L) strictly increasing in K and L. The conditional factor demands K*(Q, w, r) and L*(Q, w, r) would definitely change if w doubles and r triples.

I’m inclined to say this is false, because I can imagine a scenario where a firm far prefers labor to capital, so the amount of labor employed stays the same after the change, but I’m not sure how to find a function with those properties.

$\endgroup$
0
$\begingroup$

Consider $F(K, L) = \min(2L + K, L + 2K)$. Check that conditional factor demands satisfy :

$K^*(Q = 9, w = 1, r = 1) = L^*(Q = 9, w = 1, r = 1) = 3$ and $K^*(Q = 9, w = 2, r = 3) = L^*(Q = 9, w = 2, r = 3) = 3$

Therefore, the statement is false.

$\endgroup$
  • $\begingroup$ Thank you very much! $\endgroup$ – taurus Mar 15 at 7:16
  • $\begingroup$ I’m having a bit of trouble solving for K* and L* since MPL/MPK is constant. Can you provide any guidance? $\endgroup$ – taurus Mar 15 at 7:35
  • $\begingroup$ Taurus, if $\frac{MP_L}{MP_K}$ is constant, then you are dealing with the perfect substitutes case, in which the optimal bundle will always contain either only labor or only capital. You need to consider two cases, when isocost lines are steeper than isoquants ($\frac{MP_L}{MP_K}<\frac{w}{r}$) and when isoquants are steeper than the isocost lines ($\frac{MP_L}{MP_K}>\frac{w}{r}$). In the former case, the cost-minimizing input bundle will contain only capital, while in the latter it will contain only labor. $\endgroup$ – dlnB Mar 15 at 19:39
  • $\begingroup$ If we are in case 1 ($\frac{MP_L}{MP_K}<\frac{w}{r}$), we immediately know $L^*=0$. Plug this into the production function will allow you to solve for $K^*$. In case 2, ($\frac{MP_L}{MP_K}>\frac{w}{r}$), we immediately know $K^*=0$. Plug this into the production function will allow you to solve for $L^*$. It helps to draw the isoquant map to better understand this. $\endgroup$ – dlnB Mar 15 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.