Question

The total production\(P\)of a certain product depends on the amount\(L\)of labor used and the amount\(K\)of capital investment. The Cobb-Douglas model for the production function is\(P = b{L^\alpha }{K^{1 - \alpha }}\), where\(b\) and\(\alpha \)are positive constants and\(\alpha < 1\). 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\(L = \frac{{\alpha p}}{m}\)and \(K = \frac{{(1 - \alpha )p}}{n}\).

Short answer

So, \(L = \frac{{\alpha p}}{m}\), and \(K = \frac{{(1 - \alpha )p}}{n}\).

Step-by-step solution

Method of Lagrange multipliers

To find the maximum and minimum values of\(f(x,y,z)\)subject to the constraint\(g(x,y,z) = k\)(assuming that these extreme values exist and\(\nabla g \ne {\bf{0}}\)on the surface\(g(x,y,z) = k)\):

(a) Find all values of\(x,y,z\), and\(\lambda \)such that\(\nabla f(x,y,z) = \lambda \nabla g(x,y,z)g(x,y,z) = k\)

And

(b) Evaluate\(f\)at all the points\((x,y,z)\)that result from step (a). The largest of these values is the maximum value of\(f\); the smallest is the minimum value of\(f\).

Assumption

Given function is\(P = b{L^\alpha }{K^{1 - \alpha }}\)

Constraint is\(g(L,K) = mL + nK = p\)

Using Lagrange's method of multipliers we have:

\(\vec \nabla P = \lambda \vec \nabla g\)

Form equation

Substituting function & comparing both sides we have:

\(\begin{array}{l}b\alpha {L^{\alpha - 1}}{K^{1 - \alpha }} = \lambda m\,\,\,....(1)\\b(1 - \alpha ){L^\alpha }{K^{ - \alpha }} = \lambda n\,\,\,....(2)\\mL + nK = p\,\,\,...(3)\end{array}\)

When\(\lambda \ne 0\).

Multiply equation (1) by\(L\)we have:

\(b\alpha {L^\alpha }{K^{1 - \alpha }} = \lambda mL\,\,\,\,...(4)\)

Multiply equation (2) by\(k\)we have:

\(b(1 - \alpha ){L^\alpha }{K^{1 - \alpha }} = \lambda nK\,\,\,...(5)\)

Solve equation

Add equation (4) & (5) we have:

\({L^\alpha }{K^{1 - \alpha }}(b\alpha + b - b\alpha ) = \lambda (mL + nK)\)

\(b{L^\alpha }{K^{1 - \alpha }} = \lambda (mL + nK)\)

By equation (3); \(b{L^\alpha }{K^{1 - \alpha }} = \lambda p\,\,\,\,\,...(6)\)

By equation (2); \(\lambda = \frac{{b(1 - \alpha ){L^\alpha }{K^{ - \alpha }}}}{n}\,\,\,\,\,...(7)\)

Find value of variable

Substitute equation (7) in (6) we have:

Then \(b{L^\alpha }{K^{1 - \alpha }} = \frac{{b(1 - \alpha ){L^\alpha }{K^{ - \alpha }}p}}{n}\)

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

And\(L = \frac{{p - nK}}{m}\)

\(\begin{array}{c}L = \frac{{p - p + \alpha p}}{m}\\ = \frac{{\alpha p}}{m}\end{array}\)

Extreme points

If\(\lambda = 0\)take\(L = 0\)

Then from equation (3) \(K = \frac{p}{n}\)

Take\(K = 0\)then\(L = \frac{p}{m}\)

Then all the extreme points are\(\left( {\frac{{\alpha p}}{m},\frac{{(1 - \alpha )p}}{n}} \right)\),\(\left( {0,\frac{p}{n}} \right),\left( {\frac{p}{m},0} \right)\)

Evaluate function on extreme points

Evaluating\(P\)on these extreme points

\(P\left( {\frac{{\alpha p}}{m},\frac{{(1 - \alpha )p}}{n}} \right) = b{\left( {\frac{{\alpha p}}{m}} \right)^\alpha }{\left( {\frac{{(1 - \alpha )p}}{n}} \right)^{1 - \alpha }}\)

\(P\left( {0,\frac{p}{n}} \right) = P\left( {\frac{p}{m},0} \right) = 0\)

Therefore maximum value of \(P\)is achieved when\(L = \frac{{\alpha p}}{m}\), and \(K = \frac{{(1 - \alpha )p}}{n}\).

Hence prove.