SOLUTION: The cost function for a certain company is C = 50x + 600 and the revenue is given by R = 100x − 0.5x^2. Recall that profit is revenue minus cost. Set up a quadratic equ

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: The cost function for a certain company is C = 50x + 600 and the revenue is given by R = 100x − 0.5x^2. Recall that profit is revenue minus cost. Set up a quadratic equ      Log On


   

Question 1071397: The cost function for a certain company is
C = 50x + 600
and the revenue is given by
R = 100x − 0.5x^2.
Recall that profit is revenue minus cost. Set up a quadratic equation and find two values of x (production level) that will create a profit of $600.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
c = 50x + 600
r = 100x - .5x^2

p = r - c = 100x - .5x^2 - (50x + 600)

simplify to get:

p = 100x - .5x^2 - 50x - 600

combine like terms to get:

p = 50x - .5x^2 - 600

reorder the terms in descending order of degree to get:

p = -.5x^2 + 50x - 600

when p = 600, this equation becomes:

600 = -.5x^2 + 50x - 600

subtract 600 from both sides of this equation to get:

0 = .5x^2 + 50x - 1200

multiply both sides of this equation by -2 to get:

0 = x^2 - 100x + 2400

factor this equation to get:

(x-40) * (x-60) = 0

solve for x to get x = 40 and x = 60.

the profit will be 600 when x = 40 and when x = 600

to confirm, go back to the original equation of p = -.5x^2 + 50x - 600

when x = 40, this equation becomes:
p = -.5*40^2 + 50*40 - 600 which becomes:
p = -800 + 2000 - 600 which becomes:
p = 2000 - 1400 which becomes:
p = 600

when x = 60, this equation becomes:
p = -.5*60^2 + 50*60 - 600 which becomes:
p = -1800 + 3000 - 600 which becomes:
p = 3000 - 2400 which becomes:
p = 600

you can also solve this graphically by graphing y = -.5x^2 + 50x - 1200 and also graphing y = 600 and looking for the intersection points.

your graph will look like this:

$$$