Question

How much should the student center charge for a cup of coffee in order to maximize revenue?

A student center sells 1600 cups of coffee per day at the price of $2.40 per cup.
A market survey shows that for every $0.05 reduction in price per cup, 50 more cups of coffee will be sold.

Answer 1

`2$`

Explanation:

Let the charge per cup of coffee be `x$` . Then, the number of cups that will be sold is

`1600 +50 times (2.40-x)/0.05 = 1600 +(2400-1000x)=4000-1000x`

Thus, the revenue will be `x(4000-1000x) $= 1000x(4-x)$` . In order to maximize revenue, we need to maximize the quantity
`x(4-x)`. While one can use calculus and equate the first derivative to zero to solve the problem, I personally favor the algebraic approach:

`x(4-x) = -(x^2-4x) = 4-(x^2-4x+4) =4-(x-2)^2`

Since a perfect square can never be negative, the largest value `x(4-x)` can attain is 4, and this value is attained for `x=2`.

At the rate of `2$`, the student center will sell 2000 cups of coffee, maximizing its revenue at `4000$` up `160$` from the revenue it is posting now.