Maximizing revenue word problem?

22.November, 2009

Suppose that the management of the Ajax Dryer Company has determined that when the unit price is x dollars, the revenue R in dollars is R(x)= -4x^2+2400x.
What unit price should be established to maximize the revenue?

A. 200
B.250
C.300
D.350
E.400

We know this is a parabola so if we find the x value at the peak we know it is the maximum.

We also know that maximum occurs when the slope of the parabola is zero, so we can set the first dirivative of R(x) to zero and solve for x.

R(x)= -4x^2+2400x
R’(x) = -8x +2400
Set to zero
-8x +2400 = 0
x = 300

The unit price should be $300 for maximum revenue.

Posted by: admin in: revenue management 1 Comment »

1 Comment für “Maximizing revenue word problem?”

  1. Sassafras sagt:
    23.November, 2009 um 5:25 am

    We know this is a parabola so if we find the x value at the peak we know it is the maximum.

    We also know that maximum occurs when the slope of the parabola is zero, so we can set the first dirivative of R(x) to zero and solve for x.

    R(x)= -4x^2+2400x
    R’(x) = -8x +2400
    Set to zero
    -8x +2400 = 0
    x = 300

    The unit price should be $300 for maximum revenue.
    References :

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