Given the Production Function Q = 72X + 15X2 - X3, where Q =Output and X=Input a) What is the Marginal Product (MP) when X = 8?

b) What is the Average Product (AP) when X = 6?

c) At what value of X will Q be at its maximum?

d) At what value of X will Diminishing Returns set in?​

A. 120

B. 126

C. x=12

D. x=5

Explanation:

A. Calculation to determine the Marginal Product (MP) when X = 8

Production function=Q=72x+15x^2-x^3

Let x be 8

Marginal Product (MP)=dQ/dx

Marginal Product (MP)=72x+15x^2-x^3

Marginal Product (MP)=72+30x-3x^2

Now let plug in X = 8

Marginal Product (MP)=72+30(8)-3(8)^2

Marginal Product (MP)=72+240-192

Marginal Product (MP)=120

Therefore the Marginal Product (MP) when X = 8 will be 120

B. Calculation to determine the Average Product (AP) when X = 6

Average Product (AP)=Q/X

Average Product (AP)=72+15x-x^2

Let plug in x=6

Average Product (AP)=72+15(6)-(6)^2

Average Product (AP)=72+90-36

Average Product (AP)=126

Therefore the Average Product (AP) when X = 6 is 126

C. Calculation to determine at what value of X will Q be at its maximum

Maximizing Q=dQ/dx=0

Maximizing Q=72+30x-3x^2=0

Maximizing Q=3x^2-30x-72=0

Maximizing Q=x^2-10x-24=0

Maximizing Q=x^2-12x+2x-24=0

Maximizing Q=x(x-12)+2(x-12)=0

Hence:

x=12 or (x=-2)

Therefore at what value of X will Q be at its maximum will be at x=12

D. Calculation to determine At what value of X will Diminishing Returns set in

Diminishing returns=dMP/dx=d²Q/dx²

Diminishing returns=30-6x<0

Hence:

x=30/6<0

x=5<0

Therefore at what value of X will Diminishing Returns set in will be at x= 5 or when MP is at a MAXIMUM VALUE.

b is the correct answer

answer/// product differentiation;