Question

Suppose we have a production function for wheat 

 Q = 𝐾𝐿 − 0.8K2 − 0.2𝐿

2

A. Given K = 10 Units, calculate the level of labour that the AP reaches max and how 

much wheat is produced at that point

B. Given K = 10 at what level of labour output reaches maximum

C. How much level of labor should the company hire in order to be efficient

Answer 1

Given,

Production function:

$Q=KL−0.8K^ 2 −0.2L ^2$

(A).

$Q=10L−0.8(10)^ 2 −0.2L^ 2$

$Q=10L−0.8×100−0.2L ^2$

$Q=10L−80−0.2L^ 2$

$AP=\frac{Q }{L}$

$AP=\frac{10L−80−0.2L^ 2}{ L}$

$AP=10−\frac{80 }{ L} −0.2L$

AP will be maximum if the first order derivative is zero

$\frac{dAP }{dL} =−(\frac{−80}{ L ^2 })−0.2$

$\frac{dAP}{ dL} =\frac{80}{ L ^2} −0.2$

$\frac{dAP }{dL} =0$

$\frac{80} {L^ 2 } −0.2=0$

$\frac{80 }{ L^ 2 }=0.2$

$L^ 2 =\frac{80}{ 0.2 }$

$L^ 2 =400$

$L=20$

L=20 will give maximum AP.

Substitute K=10 in the production function.

The quantity of wheat at L=20 is given below:

$Q=10×20−80−0.2×(20)^ 2$

$Q=200−80−80$

$Q=40$

40 units of wheat will be produced.

(B).

To find maximum output we will find the first-order derivative (slope) of the production function and set that equal to zero.

$\frac{dQ }{dL} =\frac{d(10L−80−0.2L^ 2 ) }{dL}$

$\frac{dQ }{dL }=10−0.2×2L$

$\frac{dQ }{dL }=10−0.4L$

$\frac{dQ }{dL }=0$

$10−0.4L=0$

$10=0.4L$

$L=\frac{10 }{0.4}$

$L=25$

At L=25 the output will be maximum.

(c).

Since L=25 gives the maximum output the firm will hire 25 units of labor to be efficient. After L=25 the total production will decrease and it will be inefficient for the firm to produce above L=25.