joe buys mugs for $2.50 each. If he sells them for $4, he can sell 60 mugs a week. For every 50 cents increase in price, he sells 6 fewer mugs per week. Determine the price which will maximize Joe's profits.

Answer 1

Here is each case, with the values in this order: Price, mugs sold, profit per mug, Total profit

4 -- 60 -- 1.50 -- 60 x 1.5 = 90
4.5 -- 54 -- 2 -- 54 x 2 = 108
5 -- 48 -- 2.5 -- 48 x 2.5 = 120
5.5 -- 42 -- 3 -- 42 x 3 = 126
6 -- 36 -- 3.5 -- 36 x 3.5 = 126
6.5 -- 30 -- 4 -- 30 x 4 = 120
And then it begins decreasing

So, both 5.5 and 6 per mug give a maximum amount, assuming that the price must be a multiple of 50 cents

Answer 2

let the number of 50 cent increases be n

Now:
number sold = 60
price per mug = 4

after increase:
number sold = 60 - 6n
selling price = 400 + 50n

Profit = Revenue - cost
= (60-6n)(400+50n) - 250(60-6n)
= (60-6n)(400+50n - 250)
= (60-6n)(150 + 50n)
= - 300n^2 + 2100n + 9000

if you know Calculus,
d(profit) = -600n + 2100
= 0 for a max of profit
600n = 2100
n = 3.5
was expecting a whole number, but we can interpret this as
3.5 increases of 50 cents or an increase of $1.75

He should sell them at $5.75

If you don't know Calculus, find the vertex of the corresponding
quadratic function and it has to come out to the same answer