Chapter 3.5, Problem 35AYU

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

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

(a) At what prices $p$ is revenue zero?

(b) For what range of prices will revenue exceed $\$800,000$?

To determine

To find:

a. The price $p$ for which the revenue is zero.

b. The range of prices for which the revenue will exceed $\$800,000$.

Answer to Problem 35AYU

Solution:

a. For price $\$1000$, the revenue becomes zero.

b. Range of price $p$ is: $100\sqrt{5}\left(\sqrt{5}-1\right)<p<100\sqrt{5}\left(\sqrt{5}+1\right)$.

Explanation of Solution

Given:

The revenue in dollars is given by:

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

Calculation:

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

To find the price $p$ for which the revenue becomes zero is found by solving the equation:

$R\left(p\right)=-4p^2+4000p=0$

$4p^2-4000p=0$

$4p\left(p-1000\right)=0$

$p=1000$

Thus for the price $\$1000$, the revenue becomes zero.

b. To get the range of prices for which the revenue will exceed $\$800,000$:

$R\left(p\right)=-4p^2+4000p>800000$

$4p^2-4000p+800000<0$

$4\left(p^2-1000p+200000\right)<0$

$p^2-1000p+200000<0$

$\left[p-\left(500+100\sqrt{5}\right)\right]\left[p-\left(500-100\sqrt{5}\right)\right]<0$

$500-100\sqrt{5}<p<500+100\sqrt{5}$

$100\sqrt{5}\left(\sqrt{5}-1\right)<p<100\sqrt{5}\left(\sqrt{5}+1\right)$

This is the range of price $p$.