Question 1194915: A coffee shop currently sells 300 lattes a day at $3.50 each. They recently tried raising the by price by $0.25 a latte, and found that they sold 40 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 $
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x is number of steps of change
(300-40x)(3.50+0.25x)
=1050-140x+75x-10x^2
or x, the number of steps, is -10x^2-65x+1050.
The maximum for this is when x=-b/2a or 65/-20 or -3.25 steps, so the price should be dropped.
This can be checked
300*$3.50=$1050 right now
260*$3.75=975 raising
340*3.25=$1105.
The price should be decreased 3.25 *$0.25=$0.8125 to $2.6875
the number will then increase to 3.25*40 or 130 more or 430 lattes
this maximum revenue is $1155.625 or $1155.63 Ordered pair is (430, 2.6875)
Can check this by looking at 420 lattes which would be (420, 2.75) and revenue of $1102.50
and the other direction with 440 lattes and price $2.625 and revenue of $1155.
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The function N(p) is linear. If p is 0, this represent 14 25 cent steps more than the current 300 lattes, or 560 more people. The y-intercept (N) is 860 and the slope is 0.25/-40 or -0.0625 or in dollars 1/-160
N=860-160p, where p is in dollars. At a price of 860/160 or $5.375, no lattes would be sold.
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Revenue R(p)=Number * price=(860-160p)*p
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