Question

A sports company has the following production function for a certain product, where p is the number of units produced with x units of labor and y units of capital.

p(x,y)=2500x1/5y1/5
Find:
1. Number of units produced with 26 units of labor and 1333 units of capital.
2. Marginal productivities.
3. Evaluate the marginal productivities at x=25, and y=1333

Answer 1

(a) 20226 units

(b) Marginal productivities

`P_x =2500x^{-(4)/(5)} & y^(1)/(5)`

`P_y =2500 x^(1)/(5) y^{-(4)/(5)}`

(c) Evaluation of the marginal productivities

`P_x =803`

`P_y = 15`

Explanation:

Given

`P(x,y) = 2500x^(1)/(5)y^(1)/(5)`

Solving (a): P(x,y) when x = 26 and y = 1333

`P(x,y) = 2500x^(1)/(5)y^(1)/(5)` becomes

`P(26,1333) = 2500*26^(1)/(5)*1333^(1)/(5)`

`P(26,1333) = 20226` --- approximated

Solving (b): The marginal productivities

To do this, we simply calculate Px and Py

Differentiate x to give Px, so we have:

`P(x,y) = 2500x^(1)/(5)y^(1)/(5)` becomes

`P_x =2500 * x^{(1)/(5)-1} & y^(1)/(5)`

`P_x =2500 * x^{-(4)/(5)} & y^(1)/(5)`

`P_x =2500x^{-(4)/(5)} & y^(1)/(5)`

Differentiate y to give Py, so we have:

`P(x,y) = 2500x^(1)/(5)y^(1)/(5)` becomes

`P_y =2500 * x^(1)/(5) & y^{(1)/(5)-1}`

`P_y =2500 x^(1)/(5) y^{-(4)/(5)}`

Solving (c): Px and Py when x = 25 and y = 1333

`P_x =2500x^{-(4)/(5)} & y^(1)/(5)` becomes

`P_x =2500 * 25^{-(4)/(5)} * 1333^(1)/(5)`

`P_x =803` --- approximated

`P_y =2500 x^(1)/(5) y^{-(4)/(5)}` becomes

`P_y =2500 * 25^(1)/(5) * 1333^(-4)/(5)`

`P_y = 15`