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.
