Chapter 3, Problem 39RE

(a)

To determine

To express: the revenue $R$ as a function of $x$ .

Answer to Problem 39RE

The expression of the revenue $R$ is $R=\frac{-1}{10}{x}^2+150x$

Explanation of Solution

Given:

  $p=-\frac{1}{10}x+150 0\le x\le 1500$

Calculation:

By the law of economics, the revenue $R$ is given by $R=xp$ .

Thus,

  $\begin{array}{l}R=x\left(-\frac{1}{10}x+150\right)\\ =\frac{-1}{10}{x}^2+150x\end{array}$

Therefore, $R=\frac{-1}{10}{x}^2+150x$

Conclusion:

The expression of the revenue $R$ is $R=\frac{-1}{10}{x}^2+150x$

(b)

To determine

To find:the revenue if 100units are sold

Answer to Problem 39RE

The revenue if $100$ units are sold is $\$14000$

Explanation of Solution

Calculation:

Substitute $100$ for $x$ in $R=\frac{-1}{10}{x}^2+150x$ .

  $\begin{array}{l}R=\frac{-1}{10}{\left(100\right)}^2+150\left(100\right)\\ =\frac{-1}{10}\left(1000\right)+15000\\ =-1000+15000\\ =\$14000\end{array}$

Therefore, the revenue if $100$ units is sold is $\$14000$ .

Conclusion:The revenue if $100$ units is sold is $\$14000$

(c)

To determine

To find: the maximum revenue and the quantity $x$ that maximizes the revenue

Answer to Problem 39RE

When $x=750$ there will be maximum revenue

The maximum revenue will be $\$56,250$

Explanation of Solution

Calculation:

The function of $R$ is a quadratic equation with $a=\frac{-1}{10}and\text{ b}=150$ . Since $a<0$ the vertex is the highest point on the parabola.

So, the revenue will be maximum when the quantity sold $x=\frac{-b}{2a}$ .

  $x=\frac{-150}{2\left(\frac{-1}{10}\right)}=\frac{150}{\frac{1}{5}}=750$

Therefore, when $x=750$ there will be maximum revenue.

Find the maximum revenue by substituting 750 for $x$ in $R=\frac{-1}{10}{x}^2+150x$ .

  $\begin{array}{l}R=\frac{-1}{10}{\left(750\right)}^2+150\left(750\right)\\ =\frac{-1}{10}\left(562,500\right)+112,500\\ =-56,250+11,2500\\ =\$56,250\end{array}$

Therefore, the maximum revenue will be $\$56,250$ .

Conclusion:

Therefore, there will be maximum revenuewhen $x=750$

The maximum revenue will be $\$56,250$

(d)

To determine

To find:price charged by the company to maximize revenue?

Answer to Problem 39RE

  $\$75$ should be the price to maximize the revenue

Explanation of Solution

Calculation:

Substitute $750$ for $x$ in $p=-\frac{1}{10}x+150$ to get the price to maximize the revenue.

  $\begin{array}{l}p=-\frac{1}{10}\left(750\right)+150\\ =-75+150\\ =\$75\end{array}$

Therefore, $\$75$ should be the price to maximize the revenue.

Conclusion:

  $\$75$ should be the price to maximize the revenue