Optimization?

A manufacturer determines that x employees on a certain production line will produce y units per month where y=75x^2-0.2x^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.

1 Answer
Jul 16, 2017

y' = 150x - 0.8x^3

y' is never undefined

y' = 150x-0.8x^3 = 0

1500x-8x^3 = 0

4x(375-2x^2) = 0

x=0 or x = +-sqrt(375/2)

We are not interested in negative numbers of workers, so the critical numbers are

0 and sqrt(375/2)

Note that when x = 0, the production is y =0.

So we will restrict our attention again to (0,oo).

Testing to the left and right of sqrt(375/2) = sqrt(187.5) we find that this is the location of a relative maximum.

(I used 10 and 100 as my test numbers.)

With only one critical number in the interval, we can change the word "relative" to "absolute".

If we are willing to accept approximate answers , we should use sqrt187.5 workers.

If we require a whole number answer . then we need to check the whole numbers closest to the critical number.
The two closest whole numbers are 13 (whose square is 169) and 14 (whose square is 13^2 + 2(13)+1 = 169+27 = 196)

Checking the y values we find that y(14) > y(13), so we should use 14 workers.