0

Please give me any advices how to solve this. I'm good in maths, but I just don't understand these economical issues

A firm's profit is given by the function: $PROFIT(L,K) = 50 L^{0.2} K^{0.8}$, where $L$ denotes labor costs and $K$ denotes capital.

Find the level of $L$ and $K$ to maximize profit under the constrain: $L+K=1000$.

buruzaemon
  • 129
Michael
  • 33

2 Answers2

5

Use Lagrange Multipliers.

$\mathcal{L}=50L^{0.2}K^{0.8}-\lambda (1000-L-K)$

  • Take first order condition with respect to K and set it to zero:

$50\times0.8\times L^{0.2}K^{-0.2}=-\lambda$

  • Take first order condition with respect to L and set it to zero:

$50\times0.2\times L^{-0.8}K^{0.8}=-\lambda$

  • Take first order condition with respect to $\lambda$ and set it to zero:

$1000=L+K$

From the first two equations, we have $50\times0.8\times L^{0.2}K^{-0.2}=50\times0.2\times L^{-0.8}K^{0.8}$

$\Rightarrow 4L=K$

Plug this into the third equation, we have $4L+L=1000 \Rightarrow L=200$

And therefore $K=1000-L=1000-200=800$.

You will need $L=200$ and $K=800$ to maximize the profit.

George
  • 171
0

You can also do it without Lagrange multipliers since the problem is rather simple (only two variables and one linear constraint).

Extract $K$ from the constraint; so $K=1000-L$ which makes the profit to be $$P=50 (1000-L)^{4/5} L^{1/5}$$ Notice that maximizing $P$ is the same as maximizing $Q=\frac{P^5}{50}$ which is $$Q=(1000-L)^4 L$$ Now, differentiate with respect to $L$ $$\frac{dQ}{dL}=(1000-L)^4-4 (1000-L)^3 L=-5 (1000 - L)^3 (L-200 )$$ which, beside $L=1000$ of no interest (profit would be $0$), cancels if $L=200$.

The second derivative $$\frac{d^2Q}{dL^2}=20 (L-1000)^2 (L-400)$$ being negaitve if $L=200$ confirms that this point is a maximum.

Claude Leibovici
  • 260,315