Chapter 3.5, Problem 36AYU

(a)

To determine

To Calculate: The price value at which the revenue is zero.

Answer to Problem 36AYU

The price value at which the revenue is zerois $\$3800$ .

Explanation of Solution

Given Information:

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

Calculation:

The price value at which the revenue is zero is calculated as:

  $\begin{array}{l}R\left(p\right)=-\frac{1}{2}{p}^2+1900p\\ 0=-\frac{1}{2}{p}^2+1900p\\ -\frac{1}{2}p\left(p-3800\right)=0\\ p=\$3800\end{array}$

(b)

To determine

To Calculate: The price range value at which the revenue exceeds $\$1,200,000$ .

Answer to Problem 36AYU

The price range value at which the revenue exceeds $\$1,200,000$ is $\$800$ and $\$3000$ .

Explanation of Solution

Given Information:

  $-\frac{1}{2}{p}^2+1900p>\$1,200,000$

Calculation:

The price range value at which the revenue exceeds $\$1,200,000$ is calculated as:

  $\begin{array}{l}-\frac{1}{2}{p}^2+1900p>\$1,200,000\\ \frac{1}{2}{p}^2-1900p+\$1,200,000<0\\ \frac{1}{2}\left(p^2-3800p+\$2,400,000\right)<0\\ p^2-3800p+\$2,400,000<0\\ \left(p-800\right)\left(p-3000\right)<0\\ p=800,3000\end{array}$