Question

Referring to Exercise 21, we now suppose that the production is fixed at\(b{L^\alpha }{K^{1 - \alpha }} = Q\), where\(Q\)is a constant. What values of\(L\)and\(K\)minimize the cost function\(C(L,K) = mL + nK\)?

Short answer

So, the value is\(L = k = {\left( {\frac{Q}{b}} \right)^{\frac{1}{{1 - \alpha }}}}\)

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\(f\).

Apply Lagrange’s method

Consider\(C(L,k) = mL + nk\)and\(g(L,k) = b{L^\alpha }{k^{1 - \alpha }} = Q\)

Using Lagrange's method we have\(L,k\)and\(\lambda \)such that

\(\begin{array}{c}\vec \nabla C(L,k) = \lambda \vec \nabla g(L,k)\\\nabla (mL + nK) = \lambda \nabla (b{L^\alpha }{k^{1 - \alpha }})\end{array}\)

Differentiating & comparing both sides we have:

\(\begin{array}{c}m = \lambda \alpha b{L^{\alpha - 1}}{

k^{1 - \alpha }}\,\,\,\,....(1)\\n = \lambda b(1 - \alpha ){L^\alpha }{k^{ - \alpha }}\,\,\,...(2)\\b{L^\alpha }{k^{1\alpha }} = Q\,\,\,\,...(3)\end{array}\)

Solve equation

By equation (1); \(mL = \lambda \alpha b{L^\alpha }{k^{1 - \alpha }}\,\,\,\,...(4)\)

And from equation (2); \(nk = \lambda b(1 - \alpha ){L^\alpha }{k^{1 - \alpha }}\,\,\,\,\,...(5)\)

Substituting value of\(\lambda b{L^\alpha }{k^{1 - \alpha }} = \frac{{nk}}{{(1 - \alpha )}}\)from equation (5) in (4) gives

\(L = \frac{{n\alpha }}{{m(1 - \alpha )}}k\,\,\,\,....(6)\)

Using this value of\(L\)in equation (3)

\(\begin{array}{c}b{\left( {\frac{{n\alpha }}{{m(1 - \alpha )}}} \right)^\alpha } \cdot {k^{1 - \alpha }} = Q\\{k^{1 - \alpha }} = {\left( {\frac{{m(1 - \alpha )}}{{n\alpha }}} \right)^\alpha } \cdot \frac{Q}{b}\\k = {\left( {\frac{{m(1 - \alpha )}}{{n\alpha }}} \right)^{\frac{\alpha }{{1 - \alpha }}}} \cdot {\left( {\frac{Q}{b}} \right)^{\frac{1}{{1 - \alpha }}}}\end{array}\)

Find value of\({\bf{m}}\,{\bf{\& }}\,{\bf{n}}\)

Substituting value of\(k\) in equation (6) we have:

Then

\(\begin{array}{c}L = \frac{{n\alpha }}{{m(1 - \alpha )}} \cdot {\left( {\frac{{m(1 - \alpha )}}{{n\alpha }}} \right)^{\frac{\alpha }{{1 - \alpha }}}} \cdot {\left( {\frac{Q}{b}} \right)^{\frac{1}{{1 - \alpha }}}}\\ = {\left( {\frac{{m(1 - \alpha )}}{{n\alpha }}} \right)^{\frac{{2\alpha - 1}}{{1 - \alpha }}}} \cdot {\left( {\frac{Q}{b}} \right)^{\frac{1}{{1 - \alpha }}}}\end{array}\)

Also from equation (1) and (2)

\(\begin{array}{c}\frac{m}{\alpha } = \frac{n}{{1 - \alpha }}\\m(1 - \alpha ) = n\alpha \end{array}\)

This gives \(L = k = {\left( {\frac{Q}{b}} \right)^{\frac{1}{{1 - \alpha }}}}\)