A division of Ditton Industries manufactures the Futura model microwave oven. The daily cost (in dollars) of producing these mic

Question

22 days ago

7

A division of Ditton Industries manufactures the Futura model microwave oven. The daily cost (in dollars) of producing these mic

rowave 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.)

Mathematics

1 answer:

Leona [9.2K] · Answer 1 · 2026-03-05T21:56:56+03:00

Answer:

The costs associated with the production are:

For the 111st oven, it costs $114.07
The 221st oven comes to $122.71
The 311st oven amounts to $140.59

The marginal costs for x equal to 110, 220, and 310 are outlined as follows:

$C'(110)=\$114.06$

$C'(220)=\$122.64$

$C'(310)=\$140.46$

Step-by-step explanation:

The daily cost for manufacturing the microwave ovens can be expressed through the given function:

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

(a) To determine the actual cost of the 111st oven, subtract the total cost of producing the first 110 ovens from that of the first 111 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 oven:

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

For the 311st oven:

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

(b) If C(x) represents the total cost function, then the marginal cost function is its derivative, expressed as C'(x).

We need to calculate 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 costs when x equals 110, 220, and 310 are represented by

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

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

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