Question 192693This question is from textbook Intermediate Algebra
: Recall that total profit P is the difference between total revenue R and total cost C. Given R(x)=1000x-x^2 and C(x)=3000+20x, find the total profit, the maximum value of the total profit, and the value of x at which it occurs.
This question is from textbook Intermediate Algebra
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! Recall that total profit P is the difference between total revenue R and total cost C. Given R(x)=1000x-x^2 and C(x)=3000+20x, find the total profit, the maximum value of the total profit, and the value of x at which it occurs.
.
Profit = Revenue - Cost
f(x) = R(x) - C(x)
f(x) = (1000x-x^2) - (3000+20x)
f(x) = 1000x-x^2 - 3000 - 20x
f(x) = -x^2 + 980x - 3000
.
By looking at the coefficient associated with the x^2 term (-1), we know that it is a parabola that opens downward. Therefore, finding the vertex will provide the answer:
.
To do this, manipulate the equation into the "vertex form":
y= a(x-h)^2+k
.
f(x) = -x^2 + 980x - 3000
f(x) = (-x^2 + 980x) - 3000
f(x) = -(x^2 - 980x + __ ) - 3000
f(x) = -(x^2 - 980x + 240100) - 3000 + 240100
f(x) = -(x+490) + 237100
.
maximum value of the total profit: $237100
value of x at which it occurs: -490
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