Production Function (Partial Derivatives)

• Jan 14th 2012, 11:11 PM
afung22
Production Function (Partial Derivatives)
1. Consider the production function f(x,y) = 60x^3/4 y^1/4 ,which gives the number of units of goods produced when utilizing x units of labor and y units of capital.

(a) Compute ∂f/∂x and ∂f/∂y . These quantities are referred to as the ∂x/∂y
marginal productivities of labor and of capital respectively.

(b) If the amount of capital is held fixed at y = 16 and the amount of labor increases by 1 unit, estimate the increase in the quantity of goods produced.

I did part a. And the partial derivatives i got were

∂f/∂x = 45x ^ -1/4 y ^ 1/4

∂f/∂y = 15 x^ 3/4 y ^ -3/4

I got confused with part B. I know that the capital stays at y = 16 but labor increases by 1 which means x= x+1
And I use the partial derivative rule = delta P = P capital (delta capital) + P labor (delta labor)

The capital stays constant so delta capital should be 0...??

Some help would be great!
• Jan 15th 2012, 01:24 AM
CaptainBlack
Re: Production Function (Partial Derivatives)
Quote:

Originally Posted by afung22
1. Consider the production function f(x,y) = 60x^3/4 y^1/4 ,which gives the number of units of goods produced when utilizing x units of labor and y units of capital.

(a) Compute ∂f/∂x and ∂f/∂y . These quantities are referred to as the ∂x/∂y
marginal productivities of labor and of capital respectively.

(b) If the amount of capital is held fixed at y = 16 and the amount of labor increases by 1 unit, estimate the increase in the quantity of goods produced.

I did part a. And the partial derivatives i got were

∂f/∂x = 45x ^ -1/4 y ^ 1/4

∂f/∂y = 15 x^ 3/4 y ^ -3/4

I got confused with part B. I know that the capital stays at y = 16 but labor increases by 1 which means x= x+1
And I use the partial derivative rule = delta P = P capital (delta capital) + P labor (delta labor)

The capital stays constant so delta capital should be 0...??

Some help would be great!

You want:

$\displaystyle \Delta P\approx \left. \frac{\partial f}{\partial x} \right|_{y=16} \Delta x$

so as $\displaystyle \Delta x=1$

$\displaystyle \Delta P \approx \left. \frac{\partial f}{\partial x} \right|_{y=16}=45 \frac{1}{2x^{1/4}}$

CB