Optimization problem: A manufacturer determines that x employees on a certain production line will produce y units per month where y = 75 x^(2) -0.2 x^(4). To obtain the maximum…

Question

Optimization problem:

A manufacturer determines that x employees on a certain production line will
produce y units per month where
`y = 75 x^(2) -0.2 x^(4)`. To obtain the maximum
monthly production, how many employees should be assigned to the production
line?

** It is 

NOT  sufficient to find an answer that you think is a max or a min without testing for relative extrema. You  MUST  test relative extrema at all times by using either the first or second derivative test even if you only have one critical value/point.

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.