SOLUTION: Suppose that a corporation that manufactures widgets determines that its revenue function is R(x)=1000x-x^2 and its cost function is C(x)=3000+20x, where x represents the number o

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Suppose that a corporation that manufactures widgets determines that its revenue function is R(x)=1000x-x^2 and its cost function is C(x)=3000+20x, where x represents the number o      Log On


   

Question 192640: Suppose that a corporation that manufactures widgets determines that its revenue function is R(x)=1000x-x^2 and its cost function is C(x)=3000+20x, where x represents the number of widgets produced. Find the corporation's maximum profit.
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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Suppose that a corporation that manufactures widgets determines that
its revenue function is R(x)=1000x-x^2 and its cost function is C(x)=3000+20x,
where x represents the number of widgets produced.
Find the corporation's maximum profit.
:
Profit = Revenue - cost
therefore:
P = (1000x - x^2) - (3000+20x)
P = 1000x - x^2 - 3000 - 20x
P = -x^2 + 980x - 3000;
A quadratic equation, find the axis of symmetry & maximum using x = %28-b%29%2F%282a%29
In this equation: a=-1; b=980
x = %28-980%29%2F%282%2A-1%29
x = +490 units produce max profit
:
Find max profit using: P = -x^2 + 980x - 3000;
:
P = -(490^2) + 980(490) - 3000
P = -240100 + 480200 - 3000
P = $237,100 is max profit
:
You can check the math by substituting 490 for x in the original equations