Chapter 3.5, Problem 36AYU

Revenue The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price $p$, in dollars, that it charges. The revenue $R$, in dollars, is given by

$R(p)=-\frac{1}{2}{p}^2+\text{ 19}00p$

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

(b) For what range of prices will revenue exceed $\$1,200,000$?

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 $\$1,200,000$.

Answer to Problem 36AYU

Solution:

a. For price $\$3800$, the revenue becomes zero.

b. Range of price $p$ is: $800<p<3000$.

Explanation of Solution

Given:

The revenue in dollars is given by:

$R\left(p\right)=-\frac{1}{2}{p}^2+1900p$

Calculation:

a. $R\left(p\right)=-\frac{1}{2}{p}^2+1900p$

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

$R\left(p\right)=-\frac{1}{2}{p}^2+1900p=0$

$\frac{1}{2}{p}^2-1900p=0$

$\frac{1}{2}p\left(p-3800\right)=0$

$p=3800$

Thus for the price $\$3800$, the revenue becomes zero.

b. To get the range of prices for which the revenue will exceed $\$1,200,000$:

$R\left(p\right)=-\frac{1}{2}{p}^2+1900p>1200000$

$\frac{1}{2}{p}^2-1900p+1200000<0$

$\frac{1}{2}\left(p^2-3800p+2400000\right)<0$

$p^2-3800p+2400000<0$

$\left(p-800\right)\left(p-3000\right)<0$

$800<p<3000$

This is the range of price $p$.