Question

a university student sells 1600 cups of coffee per day at a price of $2.40. a market survey shows that for every $0.05 reduction in price, 50 more cups of coffee will be sold. how much should the student center charge for a cup of coffee in order to maximize revenue?

Answer 1

Revenue is quantity times price.

Answer 2

R(x)=x*p(x)

Answer 3

x = price reductions

Answer 4

Let x be the number of (5c) discounts. Lets use cents as units for revenue. Then:

Revenue = R = Volume x price = (1600+50x)(240-5x)

R = 384000 +4000 x -250 x^2

To maximise revenue dR/dx = 0

dR/dx = 4000-500x=0

x=8

I.e. 8 discounts of 0.05 cents

Answer 5

@iambatman then \(x\) isn't quantity?

Answer 6

Which means 2.40 - 8x0.05 = $ 2 for a cuppa

Answer 7

Nope

Answer 8

what is \(xp(x)\) then?

Answer 9

x = number of price reductions
the price of a cup of coffee -> p =2.4-0.05x
Number of cups of coffee sold -> 1600+50x
Revenue = R(x) = x*p(x)
= (1600+50x)(2.4-0.05)
= -2.5x^2+40x+3840

R'(x) = 40-5x = 0
5x = 40
x = 8


p = 2.40-0.05(8) = $2.00 a cup