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
Jim H
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.

Related questions
Impact of this question
3676 views around the world
You can reuse this answer
Creative Commons License