a souvenir shop sells about 300 coffee mugs per month for $7.50 each. the shop owner estimates that for each $0.75 increase in the price, he will sell about 15 fewer coffee mugs per month.

Question

a. how much should the owner change for each mug in order to maximize the monthly income from their sales?
b. What is the maximum monthly income the owner can expect to make from the mugs?

Answer 1

if there are x 75-cent increases, then there are 300-15x mugs sold.

So, now we have the revenue as

r(x) = (7.50 + 0.75x)(300-15x)
= -11.25x^2 + 112.5 + 2250

Now just find the vertex of the parabola.

Answer 2

I've never understood how to do these problems, i wonder if i could get help with this one and i'll use it as an example to solve the similar problems.

Answer 3

Just as a reminder, revenue is price * demand. That is, toe total income for all the items sold.

recall that for y = ax^2+bx+c the vertex is at x = -b/2a. So, for this one, the vertex is at

x = 112.5/22.5 = 5

Since the parabola opens downward, the vertex is a maximum. So, the maximum revenue is

r(5) = 2531.25

Don't forget all your algebra I, now that you're taking algebra II.