Question 472258: Solve the problem.
The price p and the quantity x sold of a certain product obey the demand equation
 , 0 ≤ x ≤ 320.
What price should the company charge to maximize revenue?
$80
$120
$96
$40
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the equation is:
0 <= x <= 320
the revenue equation would be r = p*x
r is the revenue
p is the price per unit
x is the number of units sold.
since p is equivalent to (-1/2)*x+160, we can replace p in the equation to get:
r = ((-1/2)*x+160)*x
this becomes:
r = (-1/2)*x^2 + 160*x
if we replace r with y, we can graph this equation.
the graph is shown below:
from the graph, it appears that revenue will peak at somewhere between 150 and 200.
we can use the max/min formula of x = -b/2a to find the x value of the max/min point.
our equation is:
y = (-1/2)*x^2 + 160*x
a = (-1/2)
b = 160
x = -b/2a becomes -160/(-1) which becomes x = 160.
this appears to be the maximum revenue point.
when x = 160, y = (-1/2)*(160)^2 + 160*160 = $12,800
when x = 150, y = (-1/2)*(150)^2 + 160*150 = $12,750
when x = 170, y = (-1/2)*(170)^2 + 160*170 = $12,750
numbers support the graph.
your answer is:
x = 160.
since p is a function of x, this makes p = (-1/2)*x + 160 which becomes:
p = (-1/2)*160 + 160 = -80 + 160 = $80.
When p = $80 and x = 160, the revenue is 80*160 = $12,800.
the company should charge $80 per unit in order to maximize revenue, based on the equations given.
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