Jk=6x, kl=3x, and jl=27
1 answer:
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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.
