Question

A division of Ditton Industries manufactures the Futura model microwave oven. The daily cost (in dollars) of producing these microwave ovens given by the following function, where x stands for the number of units produced.C(x) = 0.0002x3 - 0.06x2 + 120x + 5000(a) What is the actual cost incurred in manufacturing the 111st oven? The 221st oven? The 311st oven? (Round your answers to the nearest cent.)111st $ 221st $ 311st $ (b) What is the marginal cost when x = 110, 220, and 310? (Round your answers to the nearest cent.)

Answer 1

Answer:

The actual cost incurred in manufacturing:

• the 111st oven is $114.07
• the 221st oven is 122.71
• the 311st oven is $140.59

The marginal cost when x = 110, 220, and 310 is

$C'(110)=\ 114.06$

$C'(220)=\ 122.64$

$C'(310)=\ 140.46$

Step-by-step explanation:

We know that the daily cost function of producing microwave ovens is given by

$C(x)=0.0002x^3 - 0.06x^2 + 120x + 5000$

(a) The actual cost incurred in manufacturing the 111st oven is the difference between the total cost incurred in manufacturing the first 111 ovens and the total cost of manufacturing the first 110 ovens:

$C(111)-C(110)=[0.0002(111)^3 - 0.06(111)^2 + 120(111) + 5000]-[0.0002(110)^3 - 0.06(110)^2 + 120(110) + 5000]\\\\C(111)-C(110)=17854.27-17740.2\\\\C(111)-C(110)=114.07$

For the 221st

$C(221)-C(220)=[0.0002(221)^3 - 0.06(221)^2 + 120(221) + 5000]-[0.0002(220)^3 - 0.06(220)^2 + 120(220) + 5000]\\\\C(221)-C(220)=30748.31-30625.6\\\\C(221)-C(220)=122.71$

For the 311st

$C(311)-C(310)=[0.0002(311)^3 - 0.06(311)^2 + 120(311) + 5000]-[0.0002(310)^3 - 0.06(310)^2 + 120(310) + 5000]\\\\C(311)-C(310)=42532.79-42392.2\\\\C(311)-C(310)=140.59$

(b) If C(x) is a total cost function, then the marginal cost function is defined to be its derivative C'(x).

We need to find the derivative of the total cost function:

$\frac{d}{dx}C(x)=\frac{d}{dx}(0.0002x^3 - 0.06x^2 + 120x + 5000)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\\frac{d}{dx}C(x)=\frac{d}{dx}\left(0.0002x^3\right)-\frac{d}{dx}\left(0.06x^2\right)+\frac{d}{dx}\left(120x\right)+\frac{d}{dx}\left(5000\right)\\\\C'(x)=0.0006x^2-0.12x+120$

The marginal cost when x = 110, 220, and 310 is given by

$C'(110)=0.0006(110)^2-0.12(110)+120\\C'(110)=114.06$

$C'(220)=0.0006(220)^2-0.12(220)+120\\C'(220)=122.64$

$C'(310)=0.0006(310)^2-0.12(310)+120\\C'(310)=140.46$