Suppose that the manufacturer of a gas clothes dryer has found that, when the unit price is p dollars, the revenue R (in dollars) is R(p) = -6p^2 + 18000p. What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?
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The revenue function is a quadratic function of the form R(p) = -6p^2 + 18000p. The maximum revenue occurs at the vertex of the parabola.
The x-coordinate of the vertex of a parabola given by the equation y = ax^2 + bx + c is -b/2a.
In this case, a = -6 and b = 18000.
So, the unit price that maximizes revenue is p = -b/2a = -18000/(2*-6) = $1500.
To find the maximum revenue, we substitute p = 1500 into the revenue function:
R(1500) = -6*(1500)^2 + 18000*1500 = $13,500,000.
So, the maximum revenue is $13,500,000.
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