MATH SOLVE

3 months ago

Q:
# The Continuing Education Division at the Ozark Community College offers a total of 30 courses each semester. The courses offered are usually of two types: practical and humanistic. To satisfy the demands of the community, at least 10 courses of each type must be offered each semester. The division estimates that the revenues of offering prac- tical and humanistic courses are approximately $1500 and $1000 per course, respectively (a) Devise an optimal course offering for the college. (b) Show that the worth per additional course is $1500, which is the same as the reve- nue per practical course. What does this result mean in terms of offering additional courses?

Accepted Solution

A:

Answer:The college should offer 20 practical courses and 10 humanistic courses in order to maximize the revenue.Step-by-step explanation:a)Let x and y be x= number of practical courses the college will offery= number of humanistic courses the college will offerwe have the following inequalitiesx ≥ 10y ≥ 10x+y = 30The only points (x,y) that satisfy all this inequalities are (10,20) and (20,10) (see picture attached).On the other hand, the revenues would be given by R = 1,500x + 1,000yThe maximum of R is attained in (x,y) = (20,10) and is R = 300,000 + 100,000 = 400,000and the college should be offering 20 practical courses and 10 humanistic ones.b) If x = 21 then y must be 9, and then y does not satisfy y ≥ 10.So you can only offer additional humanistic courses.In n is an integer < 30But if y=10+n then x must be 30-n and the revenue would beR = 1,500(30-n) + 1,000(10+n) = 350,000 - 1,500n + 100,000+ 1,000n = 400,000 - 500n < 400,000This result means in terms of offering additional courses, that is not worth doing it.