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The manufacturer of a CD player has found that the revenue R (in dollars) is R(p)= -4p^2+1280p, when the unit price is p dollars. If the manufacturer sets the price p to maximize revenue, what is the maximum revenue to the nearest whole dollar?
To find the maximum revenue, we need to find the vertex of the quadratic equation.
The equation for revenue is R(p) = -4p^2 + 1280p
To find the vertex, we can use the formula: p = -b/2a
In this case, a = -4 and b = 1280.
p = -1280 / (2 * -4)
p = -1280 / -8
p = 160
Now, we can substitute this value of p into the revenue equation to find the maximum revenue:
R(p) = -4(160)^2 + 1280(160)
R(p) = -4(25600) + 204800
R(p) = -102400 + 204800
R(p) = 102400
Therefore, the maximum revenue is $102,400.
To find the maximum revenue, we need to find the value of p that maximizes the revenue function R(p) = -4p^2 + 1280p.
Step 1: Determine the critical points.
To find the critical points, we need to find the values of p where the derivative of the revenue function is zero or undefined. So, let's find the derivative of R(p) with respect to p:
R'(p) = d/dp(-4p^2 + 1280p)
= -8p + 1280
Setting R'(p) equal to zero and solving for p:
-8p + 1280 = 0
-8p = -1280
p = -1280/(-8)
p = 160
So, p = 160 is a critical point.
Step 2: Determine the nature of the critical point.
To determine if it's a maximum or minimum, we need to analyze the second derivative of R(p). We'll take the derivative of R'(p):
R''(p) = d/dp(-8p + 1280)
= -8
Since the second derivative R''(p) is a constant value (-8), and it's negative, we know that the critical point at p = 160 is a maximum.
Step 3: Find the maximum revenue.
To find the maximum revenue, substitute the value of p = 160 back into the revenue function R(p):
R(160) = -4(160)^2 + 1280(160)
= -4(25600) + 204800
= -102400 + 204800
= 102400
Therefore, the maximum revenue, to the nearest whole dollar, is $102,400.
Find the value of p that maximizes R(p).
Since you don't know calculus yet, try completing the square.
R(p) = -4[p^2 -320 p + (160)^2] + 4*(160)^2
= 25,600 - 4(p - 160)^2
Maximum revenue occurs at p = 160.