Marginal cost, maximum and minimum costs of production?

Question

A manufacturer estimates that the total cost of producing x units of its product is C(x)=4xe^(-x/6)+30 dollars.

a. Determine the marginal cost function

b. Find the maximum and minimum costs of production, if production varies between 3 and 12 units.

Good luck with this one!

Answer 1

To get part (a), use the product rule and the chain rule:
=(4)(e^(-x/6)) + (4x)[(e^(-x/6)(-1/6)]
=4e^(-x/6) - 2/3xe^(-x/6)
I factor out the e and simplify
c'(x)=e^(-x/6)(12-2x/3)

For part (b), use the marginal cost function and find where it equals zero, which is 6.

The minimum and maximum values must be one of the three values we have, which are the endpoints of the closed interval and the critical point:
C(3)=37.28
C(6)=38.83
C(12)=36.50

So, the maximum value is 6 units and the minimum value is 12 units.

Answer 2

The marginal cost function is the cost of producing the next item or C(x+1)-C(x).

So substitute.

For part b, you can also substitute. For 3 units,
C(3) = 12 e^(-1/2) + 30 = $ 38 appx
For 12 units
C(12) = 48 e^(-2) + 30 = $ 36 appx
In between:
C'(x) = 4 e^(-x/6) - 2x/3 e^(-x/6)
At the maximum or minimum C'(x)=0 , so
2x/3 = 4 and x= 6.
Substituting x=6, C(6)= 24 e^(-1) + 30 = 39 appx.
It would appear that the max cost occurs at x=6 and reaches a minimum at x=12

Answer 3

a. just differentiate: c`(x)= -4/6x^2exp(-x/6) +4exp(-x/6)
b. This will either occur at a critical point ( where marginal cost function = zero) or at one of the two boundary point.