Question

Question #c9da6

Answer 1

Integrate the difference between the revenue and cost function. `P approx $933.33` million

Explanation:

Recall that profit is equal to the difference between revenue and cost.

`P=R-C`

In this case, we have `R=150` units every single year, and `C=50+.2t^2`

Since these curves do not intersect within the domain, we can calculate the profit by finding the area beneath the revenue curve, and subtracting the area beneath the cost curve. We can do this by either integrating both functions separately and then subtracting the cost integral from the revenue integral, or (because the integral of a difference is the same as the difference of the integrals), we can integrate the difference between R and C.

Integrating with respect to t, we have:

`P = int_0^10 (R-C)dt = int_0^10 (150-50-.2t^2)dt = int_0^10 (100-0.2t^2)dt = (100(10)-(0.2/3)(10)^3) - (100(0) - (0.2/3)0^3) = 1000 - 200/3 = 3000/3-200/3 = 2800/3 approx 933.bar3`

Thus, the total profit is approximately 933.33 million dollars over 10 years.