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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 ?

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1 Answer 1

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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 $.

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  • $\begingroup$ Thank you so much for your help, however can you please explain why it is $ \lambda(Q-q) $ ? I dont understand this part? $\endgroup$
    – italy
    Jan 8, 2018 at 11:16
  • $\begingroup$ 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} $. $\endgroup$
    – kieran
    Jan 8, 2018 at 14:33
  • $\begingroup$ 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 $\endgroup$
    – italy
    Jan 8, 2018 at 15:30
  • $\begingroup$ Just edited the answer. $\endgroup$
    – kieran
    Jan 8, 2018 at 23:01

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