Question

A factory has developed a new process with savings given by S ′(t)=100−t² and increased cost C (t)=t²−10t+28. When will it no longer be profitable to use the new process?

a)t<2
b) t>2
c) t=2
d) t=5

Answer 1

The new process will cease to be profitable when the increased cost surpasses the savings, which occurs at t ≥ 9 based on the given functions. The answer options provided do not include this solution; therefore, the correct answer is none of the above.

Step-by-step explanation:

The question is asking when the new process developed by the factory will cease to be profitable. To determine this, we compare the savings function, S'(t) = 100 - t², with the increased cost function, C(t) = t² - 10t + 28. The process will no longer be profitable when the increased cost is greater than the savings. So, we need to find the value of t where S'(t) = C(t).

Setting the two equations equal to each other:

100 - t² = t² - 10t + 28

Bringing all terms to one side gives us:

0 = 2t² - 10t - 72

Dividing the entire equation by 2 gives us:

0 = t² - 5t - 36

Solving this quadratic equation by factoring, we get:

(t - 9)(t + 4) = 0

Therefore, t = 9 or t = -4. Since time cannot be negative, we discard t = -4.

The process will be profitable as long as the savings are greater than the costs, which occurs when t < 9. The process will no longer be profitable when t ≥ 9.

Therefore, none of the options provided (t < 2, t > 2, t = 2, or t = 5) are correct, and the correct answer should be t ≥ 9.