Question

an economic application: the cobb - douglas production function
a production function gives the quantity of a product that can be produced using x units of labor and y
units of capital. one common form of production function is called a cobb - douglas production function
and looks like ( p(x,y)=cx^{a}y^{1 - a} ), where ( 0lt alt1 ). notice that the exponents on labor and capital add
up to one. the values of c and a will be provided to you. for a production function ( p(x,y) ), the partial
derivative ( \frac{partial p}{partial x}=p_{x} ) is called the marginal productivity of labor (or the marginal product of labor). the
partial derivative ( \frac{partial p}{partial y}=p_{y} ) is called the marginal productivity of capital (or the marginal product of
capital).
example 3:
suppose that ( p(x,y)=6x^{0.4}y^{0.6} ) represents the number of lawns that can be maintained by a
gardening service using x workers and y mowers.

1. how many lawns can be done with 25 workers and 10 mowers?

( p(25,10)=6(25)^{0.4}(10)^{0.6}=87 ) lawns

1. if the service is currently using 25 workers and 10 mowers, find the marginal product of labor and the

marginal product of capital.

Explanation:

Step1: Find the partial derivative of \(P(x,y)\) with respect to \(x\)

Given \(P(x,y)=Cx^{a}y^{1 - a}\), using the power rule for partial derivatives \(\frac{\partial P}{\partial x}=Ca x^{a - 1}y^{1 - a}\).

Step2: Find the partial derivative of \(P(x,y)\) with respect to \(y\)

Using the power rule for partial derivatives \(\frac{\partial P}{\partial y}=C(1 - a)x^{a}y^{-a}\).

Step3: Substitute \(x = 25\), \(y = 10\), \(C = 6\), \(a=0.4\) into \(\frac{\partial P}{\partial x}\)

\(\frac{\partial P}{\partial x}=6\times0.4\times25^{0.4 - 1}\times10^{1 - 0.4}\)
First, \(25^{-0.6}=\frac{1}{25^{0.6}}=\frac{1}{(5^{2})^{0.6}}=\frac{1}{5^{1.2}}\), \(10^{0.6}=\sqrt[5]{10^{3}}\)
\(\frac{\partial P}{\partial x}=2.4\times\frac{1}{5^{1.2}}\times\sqrt[5]{10^{3}}\)
Another way: \(25^{0.4}=\sqrt[5]{25^{2}}=\sqrt[5]{625}\approx5.25\), \(10^{0.6}=\sqrt[5]{1000}\approx3.98\)
\(\frac{\partial P}{\partial x}=6\times0.4\times25^{- 0.6}\times10^{0.6}=2.4\times\frac{10^{0.6}}{25^{0.6}}=2.4\times(\frac{10}{25})^{0.6}=2.4\times(0.4)^{0.6}\approx2.4\times0.696\approx1.67\)

Step4: Substitute \(x = 25\), \(y = 10\), \(C = 6\), \(a = 0.4\) into \(\frac{\partial P}{\partial y}\)

\(\frac{\partial P}{\partial y}=6\times(1 - 0.4)\times25^{0.4}\times10^{-0.4}\)
\(25^{0.4}=\sqrt[5]{25^{2}}=\sqrt[5]{625}\approx5.25\), \(10^{-0.4}=\frac{1}{10^{0.4}}=\frac{1}{\sqrt[5]{100}}\approx0.398\)
\(\frac{\partial P}{\partial y}=3.6\times5.25\times0.398\approx3.6\times2.09\approx7.52\)

Answer:

The marginal product of labor is approximately \(1.67\) and the marginal product of capital is approximately \(7.52\)