Question

Optimization problem:

Answer 1

Cal problem!

given production 
P(x)=75x^2-0.2x^4

To find relative extrema, we need to find P'(x) and solve for P'(x)=0.

P'(x)=150x-0.8x^3    [by the power rule]

Setting P'(x)=0 and solve for extrema.
150x-0.8x^3=0  =>
x(150-0.8x^2)=0 =>
0.8x(187.5-x^2)=0
0.8x(5sqrt(15/2)-x)(5sqrt(15/2)+x)=0
=>
x={0,+5sqrt(15/2), -5sqrt(15/2)}   by the zero product rule.
[note: eqation P'(x)=0 can also be solved by the quadratic formula]

Reject negative root because we cannot hire negative persons.

So possible extrema are x={0,5sqrt(15/2)}

To find out which are relative maxima, we use the second derivative test.  Calculate P"(x), again by the power rule, 
P"(x)=-1.6x
For a relative maximum, P"(x)<0, so
P"(0)=0  which is not <0  [in fact, it is an inflection point]
P"(5sqrt(15/2))=-8sqrt(15/2) < 0, therefore x=5sqrt(15/2) is a relative maximum.

However, 5sqrt(15/2)=13.693 persons, which is impossible, so we hire either 13 or 14, but which one?

Let's go back to P(x) and find that
P(13)=6962.8
P(14)=7016.8

So we say that assigning 14 employees will give a maximum output.