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

Question

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Respuesta :

MrRoyal MrRoyal · Answer 1 · 2021-07-12T14:50:32+02:00

Answer:

(a) 20226 units

(b) Marginal productivities

[tex]P_x =2500x^{-\frac{4}{5}} & y^\frac{1}{5}[/tex]

[tex]P_y =2500 x^\frac{1}{5} y^{-\frac{4}{5}}[/tex]

(c) Evaluation of the marginal productivities

[tex]P_x =803[/tex]

[tex]P_y = 15[/tex]

Step-by-step explanation:

Given

[tex]P(x,y) = 2500x^\frac{1}{5}y^\frac{1}{5}[/tex]

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

[tex]P(x,y) = 2500x^\frac{1}{5}y^\frac{1}{5}[/tex] becomes

[tex]P(26,1333) = 2500*26^\frac{1}{5}*1333^\frac{1}{5}[/tex]

[tex]P(26,1333) = 20226[/tex] --- approximated

Solving (b): The marginal productivities

To do this, we simply calculate Px and Py

Differentiate x to give Px, so we have:

[tex]P(x,y) = 2500x^\frac{1}{5}y^\frac{1}{5}[/tex] becomes

[tex]P_x =2500 * x^{\frac{1}{5}-1} & y^\frac{1}{5}[/tex]

[tex]P_x =2500 * x^{-\frac{4}{5}} & y^\frac{1}{5}[/tex]

[tex]P_x =2500x^{-\frac{4}{5}} & y^\frac{1}{5}[/tex]

Differentiate y to give Py, so we have:

[tex]P(x,y) = 2500x^\frac{1}{5}y^\frac{1}{5}[/tex] becomes

[tex]P_y =2500 * x^\frac{1}{5} & y^{\frac{1}{5}-1}[/tex]

[tex]P_y =2500 x^\frac{1}{5} y^{-\frac{4}{5}}[/tex]

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

[tex]P_x =2500x^{-\frac{4}{5}} & y^\frac{1}{5}[/tex] becomes

[tex]P_x =2500 * 25^{-\frac{4}{5}} * 1333^\frac{1}{5}[/tex]

[tex]P_x =803[/tex] --- approximated

[tex]P_y =2500 x^\frac{1}{5} y^{-\frac{4}{5}}[/tex] becomes

[tex]P_y =2500 * 25^\frac{1}{5} * 1333^\frac{-4}{5}[/tex]

[tex]P_y = 15[/tex]