Chapter 3.3, Problem 89AYU

Maximizing Revenue 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

$R\left(p\right)=-4p^2+4000p$

What unit price should be established for the dryer to maximize revenue? What is the maximum revenue?

To determine

To calculate: The maximum revenue and the unit price that establishes the maximum revenue.

Answer to Problem 89AYU

The unit prize for the dryer to maximize the revenue is $\$500$ and the maximum revenue will be $\$1000000$.

Explanation of Solution

Given:

The revenue function, $R$ of the gas clothes dryer in unit price $p$ is given as

$R(p)=-4p^2+4000p$

Formula used:

Consider a quadratic function of the form $f(x) = a{x}^2 + bx + c, a \ne 0$.

Then the graph of the above function is a parabola with vertex $\left(\frac{-b}{2a}, f\left(\frac{-b}{2a}\right)\right)$.

This vertex is the highest point if $a < 0$ and the lowest point if $a > 0$.

Therefore, the maximum (or the minimum) value of the function will be $f\left(\frac{-b}{2a}\right)$.

Calculation:

The given function $R(p)=-4p^2+4000p$ is a quadratic function.

Thus, here, we have $a=-4<0$.

Therefore, the vertex is the maximum point in the graph of the given function.

Thus, we have

$\frac{-b}{2a}=\frac{-4000}{2(-4)}=500$.

$R\left(\frac{-b}{2a}\right)=R(500)=-4{(500)}^2+4000(500)=1000000$.

Thus, we get the vertex of the function as $(500,1000000)$.

Therefore, the unit prize for the dryer to maximize the revenue is $\$500$ and the maximum revenue is $\$1000000$.