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Anna007 [38]
3 years ago
11

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?

Mathematics
1 answer:
dybincka [34]3 years ago
8 0

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  

<em>x= number of practical courses the college will offer</em>

<em>y= number of humanistic courses the college will offer</em>

we have the following inequalities

x ≥ 10

y ≥ 10

x+y = 30

The 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 + 1,000y

The maximum of R is attained in (x,y) = (20,10) and is  

R = 300,000 + 100,000 = 400,000

and 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 < 30

But if y=10+n  then x must be 30-n and the revenue would be

R = 1,500(30-n) + 1,000(10+n) =  

350,000 - 1,500n + 100,000+ 1,000n =  

400,000 - 500n < 400,000

This result means in terms of offering additional courses, that is not worth doing it.

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