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.

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:

[tex]\frac{dP}{dL}=b\alphaL^{\alpha-1}K^{1-\alpha}[/tex]

[tex]\frac{dP}{dK}=b(1-\alpha)L^{\alpha}K^{-\alpha}[/tex]

4) Apply the method of lagrange multipliers

Using this method we have this system of equations:

[tex]\frac{dP}{dL}=\lambda m[/tex]

[tex]\frac{dP}{dK}=\lambda n[/tex]

[tex]mL+nK=p[/tex]

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

[tex]b\alphaL^{\alpha-1}K^{1-\alpha}=\lambda m[/tex] (2)

[tex]b(1-\alpha)L^{\alpha}K^{-\alpha}=\lambda n[/tex] (3)

[tex]mL+nK=p[/tex] (4)

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

[tex]\frac{\lambda m}{\lambda n}=\frac{b\alphaL^{\alpha-1}K^{1-\alpha}}{b(1-\alpha)L^{\alpha}K^{-\alpha}}[/tex] (4)

And simplyfing equation (4) we got:

[tex]\frac{m}{n}=\frac{\alpha K}{(1-\alpha)L}[/tex] (5)

4) Solve for L and K

We can cross multiply equation (5) and we got

[tex]\alpha Kn=m(1-\alpha)L[/tex]

And we can set up this last equation equal to 0

[tex]m(1-\alpha)L-\alpha Kn=0[/tex] (6)

Now we can set up the following system of equations:

[tex]mL+nK=p[/tex] (a)

[tex]m(1-\alpha)L-\alpha Kn=0[/tex] (b)

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

[tex]Lm=\alpha p[/tex]

And we can solve for L on this case:

[tex]L=\frac{\alpha p}{m}[/tex]

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

[tex]m(\frac{\alpha P}{m})+nK=p[/tex]

[tex]\alpha P +nK=P[/tex]

[tex]nK=P(1-\alpha)[/tex]

[tex]K=\frac{P(1-\alpha)}{n}[/tex]

With this we have completed the proof.