A coffee shop currently sells 420 lattes a day at $3.00 each. They recently tried raising the by price by $0.25 a latte, and found that they sold 60 less lattes a day.

a) Assume that the number of lattes they sell in a day,
N
, is linearly related to the sale price,
p
(in dollars). Find an equation for N as a function of p.
N(p)=

b) Revenue (the amount of money the store brings in before costs) can be found by multiplying the cost per cup times the number of cups sold. Again using
p
as the sales price, use your equation from above to write an equation for the revenue,
R as a function of p.
R(p) =

c) The store wants to maximize their revenue (make as much money as possible). Find the value of
p
that will maximize the revenue (round to the nearest cent).

p =

which will give a maximum revenue of $

Given that a coffee shop currently sells 420 lattes a day at $ 3.00 each, and they recently tried raising the price by $ 0.25 a latte, and found that they sold 60 less lattes a day, to determine the price that will maximize profit must be made the following calculation:

420 x 3 = 1260
360 x 3.25 = 1170
480 x 2.75 = 1320
540 x 2.5 = 1350
600 x 2.25 = 1350
660 x 2 = 1320
570 x 2.375 = 1353.75

Therefore, the price that will maximize profit is $ 2.37.

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