SOLUTION: A resturaunt sells about 330 sandwhiches each day at a price of $6 each. For each $.25 decrease in price, 15 more sandwhiches are sold per day. How much should the restauraunt char

Question 67514: A resturaunt sells about 330 sandwhiches each day at a price of $6 each. For each $.25 decrease in price, 15 more sandwhiches are sold per day. How much should the restauraunt charge to maximize daily revenue? What is the maximum daily revenue?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A restaurant sells about 330 sandwiches each day at a price of $6 each. For each $.25 decrease in price, 15 more sandwiches are sold per day. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue?
:
Let x = amt ($) decrease in price
:
Since there is an increase of 15 for each $.25 decrease, $1 decrease: 4*15 = 60
Let 60x = increase in sandwiches sold
:
Price = (6 - x)
Number sold = (330 + 60x)
:
Revenue = price * number sold: Let y = Revenue:
:
y = (6-x)(330+60x)
FOIL
y = 1980 + 360x -330x - 60x^2
:
A quadratic equation:
y = -60x^2 + 30x + 1980
:
Find the axis of symmetry: x = -b/(2a); a=-60; b=+30
x = -30/(2*-60)
x = -30/-120
x = +.25
:
Max revenue occurs at 6 - .25 = $5.75 for the sandwich
:
Find Max revenue: The problem states that for every $.25, they sell 15 more:
(330+15) * 5.75 = $1983.75 is the max revenue.
:
You can also find max revenue by substituting .25 for x in the equation:
-60(.25^2) + 30(.25) + 1980 = 1983.75
:
A graph would look like this:

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