SOLUTION: assume that i sell 66 sandwiches when selling them for $1. if i lose 5 customers each time i raise the price $.25, what is the price that will produce the higherst amount of sales?

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: assume that i sell 66 sandwiches when selling them for $1. if i lose 5 customers each time i raise the price $.25, what is the price that will produce the higherst amount of sales?      Log On


   

Question 308886: assume that i sell 66 sandwiches when selling them for $1. if i lose 5 customers each time i raise the price $.25, what is the price that will produce the higherst amount of sales? what is the amount of money i will bring in? solve using quadratics.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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assume that i sell 66 sandwiches when selling them for $1. if i lose 5 customers each time i raise the price $.25, what is the price that will
produce the highest amount of sales?
what is the amount of money i will bring in? solve using quadratics.
:
Let x = no. of .25 increases & no. of 5 customer decreases
:
Revenue = price * no. sold
R(x) = (1 + .25x)(66 - 5x)
FOIL
R(x) = 66 - 5x + 16.5x - 1.25x^2
R(x) = 66 + 11.5x - 1.25x^2
A quadratic equation
y = -1.25x^2 + 11.5x + 66
Find the axis of symmetry; x = -b/(2a), a=-1.25, b=+11.5
x = %28-11.5%29%2F%282%2A-1.25%29
x = %28-11.5%29%2F%28-2.5%29
x = 4.6 ~ 5, have to have an integer
Price for max revenue: 1 + 5(.25) = $2.25
No. sold: 66 - 5(5) = 41 sold
:
Find the max revenue; substitute 5 for x in the quadratic equation
R(x) = -1.25(5^2) + 11.5(5) + 66
R(x) = -1.25(25) + 57.5 + 66
R(x) = -32.25 + 57.5 + 66
R(x) = $92.25
:
Check: Price * no. sold
2.25 * 41 = 92.25 max sales