Chapter 4, Problem 24CR

(a)

To determine

The revenue $R$ as function of $x$ , where the price $p$ and the quantity $x$ sold of a

certain product follows the demand equation $p=-\frac{1}{10}x+150$ , where $0\le x\le 1500$

Answer to Problem 24CR

Solution:

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

Explanation of Solution

Given information:

The price $p$ and the quantity $x$ sold of a certain product follows the demand function

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

The revenue $R=xp$ , where $x$ is the quantity of a certain product sold and $p$ is the price.

Here, $p=-\frac{1}{10}x+150$ .

By substituting the value of $p$ in $R=xp$

It gives $R=x\left(-\frac{1}{10}x+150\right)$

Now, by simplifying it,

  $R=-\frac{1}{10}{x}^2+150x$

Hence, the revenue as a function of $x$ is $R=-\frac{1}{10}{x}^2+150x$

(b)

To determine

The revenue $R$ when $100$ units sold, where the price $p$ and the quantity $x$ sold of a certain product follows the demand function $p=-\frac{1}{10}x+150$ , where $0\le x\le 1500$

Answer to Problem 24CR

Solution:

The revenue when $100$ units sold is $\$14000$

Explanation of Solution

Given information:

The price $p$ and the quantity $x$ sold of a certain product follows the demand function $p=-\frac{1}{10}x+150$ , where $0\le x\le 1500$

From part (a)

The revenue $R=-\frac{1}{10}{x}^2+150x$ , where $x$ is the quantity of a certain product sold

For $x=100$

By substituting $x=100$ in the revenue function

  $R=-\frac{1}{10}{\left(100\right)}^2+150\left(100\right)$

  $\Rightarrow R=-\frac{1}{10}\left(10000\right)+15000$

  $\Rightarrow R=-1000+15000$

  $\Rightarrow R=14000$

Hence, the revenue when $100$ units sold is $\$14000$

(c)

To determine

The quantity $x$ which maximizes the revenue and the value of maximum revenue.

Answer to Problem 24CR

Solution:

The quantity $x$ which maximizes the revenue is $750$ units.

The maximum revenue is $\$56250$ .

Explanation of Solution

Given information:

The price $p$ and the quantity $x$ sold of a certain product follows the demand function $p=-\frac{1}{10}x+150$ , where $0\le x\le 1500$

From part (a),

The revenue $R=-\frac{1}{10}{x}^2+150x$ , where $x$ is the quantity of a certain product sold

Here, revenue $R$ is a quadratic function

By comparing the quadratic function $R=-\frac{1}{10}{x}^2+150x$ with general quadratic equation $a{x}^2+bx+c$

  $a=-\frac{1}{10}$ , $b=150$ and $c=0$

Since $a=-\frac{1}{10}<0$ , so the maximum value of $R$ occurs at the vertex.

The $x$ coordinate of the vertex is $x=-\frac{b}{2a}$ .

By substituting value of b and a in $-\frac{b}{2a}$

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

  $=-\frac{150\times 10}{-2}$

  $=150\times 5$

  $=750$

Thus, the $x$ coordinate of the vertex is $750$ .

So the quantity $x=750$ maximizes the revenue.

Now, to determine the maximum revenue for that substitute $x=750$ in $R$ .

  $R=-\frac{1}{10}{\left(750\right)}^2+150\left(750\right)$

  $=-\frac{1}{10}\left(562500\right)+112500$

  $=-56250+112500$

  $=-56250+112500$

  $=56250$

Thus, the maximum revenue is $\$56250$ .

(d)

To determine

The price should be charged by the company to maximize the revenue.

Answer to Problem 24CR

Solution:

The price should be charged by company to maximize the revenue is $\$75$ .

Explanation of Solution

Given information:

The price $p$ and the quantity $x$ sold of a certain product follows the demand function $p=-\frac{1}{10}x+150$ where $0\le x\le 1500$ .

From part (c),

The revenue is maximizes for the quantity $x=750$ .

By substituting $x=750$ in demand equation $p=-\frac{1}{10}x+150$

  $p=-\frac{1}{10}\left(750\right)+150$

  $=-75+150$

  $=75$

Thus, the price charged by company should be $\$75$ to maximize the revenue.