# constrained optimisation problem

Consider a firm with two inputs K and L that produce an output Q(K, L). The firm’s cost function is C(K, L) = K + 2L. It is required to minimize C(K, L) subject to the constraint Q(K, L) = q where q is a positive real constant. You may assume that the optimization problem has a solution at an interior point of R2+.

        (i) Show that


$\frac{\partial Q}{\partial L} = 2\frac{\partial Q}{\partial K}$ holds at the optimal point.

       (ii) Assume that Q(K,L) is homogeneous of degree m. At the optimal point, also


assume that ∂Q/∂K = r

for some positive real r. Find the constrained minimum value of C(K, L) in terms of m,r and q.

I am not sure how to do this problem or where to start?

would the set up be something like,

$L( c, \lambda) = K + 2L + \lambda q$

$\frac{\partial Q}{\partial L} = 2$

$\frac{\partial Q}{\partial K} = 1$

$\frac{\partial Q}{\partial \lambda} = q$

how would I find the constrained minimum value ?

Let $F = K + 2L + \lambda (Q - q)$

$\frac{\partial F}{\partial K} = 1 + \lambda \frac{\partial Q}{\partial K} = 0$

$\frac{\partial F}{\partial L} = 2 + \lambda \frac{\partial Q}{\partial L} = 0$

So, (i) is true and $\frac{\partial Q}{\partial L} = 2r$

Using (ii), Euler's Theorem states that

$K\frac{\partial Q}{\partial K} + L\frac{\partial Q}{\partial L} = mQ$

So, $Kr + 2Lr = mq$

Then, $C(K,L) = K + 2L = mq/r$.

• Thank you so much for your help, however can you please explain why it is $\lambda(Q-q)$ ? I dont understand this part? Jan 8, 2018 at 11:16
• It makes no difference if you use $\lambda (Q-q)$ or $\lambda Q$. Some authors would use $\lambda (Q-q)$ where $Q-q=0$ is the constraint. They would then have $\frac{\partial F}{\partial \lambda} = 0$ and might also calculate $\frac{\partial F}{\partial q}$. Jan 8, 2018 at 14:33
• Hi thank you for clarifying! Just one more clear up if you don’t mind. Why does $df/dL = 1 + \lambda dQ\dK$ ?? I don’t understand this part Jan 8, 2018 at 15:30
• Just edited the answer. Jan 8, 2018 at 23:01