Question

1.     Suppose that the short-run production function of certain cut-flower firm is given by: Q= 4KL -0.6K² -0.1L² where Q is quantity of cut-flower produced, L is labour input and K is fixed capital input (K=5).

 a) Determine the average product of labour (APL) function.

b) At what level of labour does the total output of cut-flower reach the maximum?

c) What will be the maximum achievable amount of cut-flower production? 

Answer 1

$Q= 4KL-0.6K^2-0.1L^2$

K= 5

Therefore by substituting K in the above function we get,

$Q= 4*5L-0.6(5)^2-0.1L^2$

$= 20L-15-0.1L^2$

By rearranging

$Q= -0. 1L^2+20L-15 .............. Equation-1$

Part (a) - Average product of labour function= $\frac{Q}{L}$

Therfore Equation $\frac{1}{L}$ ,we get,

$APL= \frac{(-0.1L^2+ 20L-15)} {L}$

$APL = -0. 1L + 20-(\frac{15}{L})$

Part(b) - Now differentiate the Equation -1 with respect to labour(L)

$\frac{dQ}{dL}=\frac{ [d(-0.1L^2+20L-15) ]}{dL}$

= -0. 2L+20

Now according to first order condition,

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

Therefore, -0. 2L+20= 0

L= 100

Part(c) - Now substitute the value of labour (L) in equation-1 to get maximum output,

We get,

$Q= -0. 1(100) ^2+20\times100-15$

Q = -1000+2000-15

Q= 985

Therefore maximum output will be 985 quantities.