SOLUTION: The cost function for a certain company is C = 40x + 800 and the revenue is given by R = 100x − 0.5x^2. Recall that profit is revenue minus cost. Set up a quadratic equation

Algebra ->  Trigonometry-basics -> SOLUTION: The cost function for a certain company is C = 40x + 800 and the revenue is given by R = 100x − 0.5x^2. Recall that profit is revenue minus cost. Set up a quadratic equation       Log On


   



Question 1129290: The cost function for a certain company is C = 40x + 800 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 $800.
x =

Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The cost function for a certain company is C = 40x + 800 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 $800.
------
Profit:: P(x) = 100x-0.5x^2 - (40x+800)
P(x) = -5x^2 +60x - 800
-----
Solve for "x"::
-5x^2+60x - 800 = 800
-5x^2 + 60x - 1600 = 0
Note:: This quadratic has no Real Number solutions.
Please check your post.
----
Cheers,
Stan H.
---------

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

if C+=+40x+%2B+800+ and the revenue R+=+100x+-0.5x%5E2
profit is P=R-C
P=100x+-0.5x%5E2-%2840x+%2B+800%29
P=100x+-0.5x%5E2-40x+-+800
P=-0.5x%5E2%2B60x+-+800
profit of P=800

800=-0.5x%5E2%2B60x+-+800

0=-0.5x%5E2%2B60x+-+800-800
-0.5x%5E2%2B60x+-+1600=0...factor
-0.5%28x%5E2-120x+%2B+3200%29=0

write -120x as -40x-80x, than we have

-0.5%28x%5E2-40x+-80x%2B3200%29=0

-0.5%28%28x%5E2-80x%29+-%2840x-+3200%29%29=0

-0.5%28x%28x-80%29+-40%28x-+80%29%29=0

-0.5+%28x+-+80%29+%28x+-+40%29+=+0
solutions:
+%28x+-+80%29+=+0=>x=80
+%28x+-+40%29+=+0=>x=40