Question

Let production be given by P = bLαK1−α where b and α are positive and α < 1. If the cost of a unit of labor is m and the cost of a unit of capital is n, and the company can spend only p dollars as its total budget, then maximizing the production P is subject to the constraint mL + nK = p. Show that the maximum production occurs when L=αp/m and K=(1-α)p/n.

Answer 1

Answer:

The proof is completed below

Step-by-step explanation:

1) Definition of info given

We have the function that we want to maximize given by (1)

$P(L,K)=bL^{\alpha}K^{1-\alpha}$   (1)

And the constraint is given by $mL+nK=p$

2) Methodology to solve the problem

On this case in order to maximize the function on equation (1) we need to calculate the partial derivates respect to L and K, since we have two variables.

Then we can use the method of Lagrange multipliers and solve a system of equations. Since that is the appropiate method when we want to maximize a function with more than 1 variable.

The final step will be obtain the values K and L that maximizes the function

3) Calculate the partial derivates

Computing the derivates respect to L and K produce this:

$\frac{dP}{dL}=b\alphaL^{\alpha-1}K^{1-\alpha}$

$\frac{dP}{dK}=b(1-\alpha)L^{\alpha}K^{-\alpha}$

4) Apply the method of lagrange multipliers

Using this method we have this system of equations:

$\frac{dP}{dL}=\lambda m$

$\frac{dP}{dK}=\lambda n$

$mL+nK=p$

And replacing what we got for the partial derivates we got:

$b\alphaL^{\alpha-1}K^{1-\alpha}=\lambda m$   (2)

$b(1-\alpha)L^{\alpha}K^{-\alpha}=\lambda n$   (3)

$mL+nK=p$   (4)

Now we can cancel the Lagrange multiplier $\lambda$ with equations (2) and (3), dividing these equations:

$\frac{\lambda m}{\lambda n}=\frac{b\alphaL^{\alpha-1}K^{1-\alpha}}{b(1-\alpha)L^{\alpha}K^{-\alpha}}$   (4)

And simplyfing equation (4) we got:

$\frac{m}{n}=\frac{\alpha K}{(1-\alpha)L}$   (5)

4) Solve for L and K

We can cross multiply equation (5) and we got

$\alpha Kn=m(1-\alpha)L$

And we can set up this last equation equal to 0

$m(1-\alpha)L-\alpha Kn=0$   (6)

Now we can set up the following system of equations:

$mL+nK=p$   (a)

$m(1-\alpha)L-\alpha Kn=0$   (b)

We can mutltiply the equation (a) by $\alpha$ on both sides and add the result to equation (b) and we got:

$Lm=\alpha p$

And we can solve for L on this case:

$L=\frac{\alpha p}{m}$

And now in order to obtain K we can replace the result obtained for L into equations (a) or (b), replacing into equation (a)

$m(\frac{\alpha P}{m})+nK=p$

$\alpha P +nK=P$

$nK=P(1-\alpha)$

$K=\frac{P(1-\alpha)}{n}$

With this we have completed the proof.