Chapter 3.4, Problem 3AYU

(a)

To determine

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

Answer to Problem 3AYU

The expression of the revenue $R$ is $R=-\frac{1}{6}{x}^2+100x$

Explanation of Solution

Given:

  $p=-\frac{1}{6}x+100$

Calculation:

The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation

  $p=-\frac{1}{6}x+100$

Find a model that expresses the revenue $R$ as a function of $x$ .

The revenue $R=px$ , using the demand equation, express $R$ as a function of $x$

  $\begin{array}{l}R=\left(-\frac{1}{6}x+100\right)x\\ =-\frac{1}{6}{x}^2+100x\end{array}$

Hence, $R=-\frac{1}{6}{x}^2+100x$

Conclusion:

Hence, $R=-\frac{1}{6}{x}^2+100x$

(b)

To determine

To find: the domain of $R$

Answer to Problem 3AYU

The domain of $R$ is $\left\{x|0\le x\le 600\right\}$ .

Explanation of Solution

Given:

  $p=-\frac{1}{6}x+100$

Calculation:

  $x$ represents the number of quantity sold, price $p>0$ , so $-\frac{1}{6}{x}^2+100x>0$ .

Solving this linear inequality, find that $x<600$ . In addition, quantity sold $x\ge 0$ . Combining these inequalities, the domain of $R$ is $\left\{x|0\le x\le 600\right\}$ .

Conclusion:

Hence, the domain of $R$ is $\left\{x|0\le x\le 600\right\}$ .

(c)

To determine

To find:the revenue if $200$ units are sold

Answer to Problem 3AYU

Hence, the revenue is $\$13,333.33$ .

Explanation of Solution

Given:

  $p=-\frac{1}{6}x+100$

Calculation:

Find the value of revenue if $200$ units are sold.

  $\begin{array}{l}R=-\frac{1}{6}{x}^2+100x\\ =-\frac{1}{6}{\left(200\right)}^2+100\left(200\right)\\ =-6666.66+20000\\ =13333.33\end{array}$

Hence, the revenue is $\$13,333.33$ .

Conclusion:

Hence, the revenue is $\$13,333.33$ .

(d)

To determine

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

Answer to Problem 3AYU

The revenue is $\$15,000$

  $300$ Units sold to maximize the revenue.

Explanation of Solution

Given:

  $p=-\frac{1}{6}x+100$

Calculation:

To find the quantity $x$ that maximizes the revenue, take the first derivative of $R$ with respect to $x$ and set it to zero.

  $\begin{array}{l}\frac{dR}{dx}=0\\ -\frac{1}{6}2x+100=0\\ -\frac{1}{3}x=-100\\ x=300\end{array}$

Hence, $300$ units sold to maximize the revenue.

He value of revenue if $300$ units are sold.

  $\begin{array}{l}R=-\frac{1}{6}{x}^2+100x\\ =-\frac{1}{6}{\left(300\right)}^2+100\left(300\right)\\ =-15000+30000\\ =15000\end{array}$

Hence, the revenue is $\$15,000$

Conclusion:

Hence, the revenue is $\$15,000$ and 300units sold to maximize the revenue.

(e)

To determine

To find:price charged by the company to maximize revenue

Answer to Problem 3AYU

Hence, company should $\$50$ to maximize the revenue.

Explanation of Solution

Given:

  $p=-\frac{1}{6}x+100$

Calculation:

Find the price should the company charge to maximize the revenue.

Since $300$ units are sold to maximize the revenue, the price is

  $\begin{array}{l}p=-\frac{1}{6}x+100\\ =-\frac{1}{6}\left(300\right)+100\\ =-50+100\\ =50\end{array}$

Hence, company should $\$50$ to maximize the revenue.

Conclusion:

Hence, company should $\$50$ to maximize the revenue.