Chapter 3.4, Problem 6AYU

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

$x=-20p+500$, $0<p\le 25$

(a) Express the revenue $R$ as a function of $x$.

(b) What is the revenue if 20 units are sold?

(c) What quantity $x$ maximizes revenue? What is the maximum revenue?

(d) What price should the company charge to maximize revenue?

(e) What price should the company charge to earn at least $\$3000$ in revenue?

To determine

To calculate:

The revenue as a function of x.

What is the revenue if 20 units are sold?

What quantity x maximises the revenue and what is the maximum revenue.

What price should the company charge to maximise the revenue?

What price should the company charge to earn at least \$3000 in revenue?

Answer to Problem 6AYU

Solution:

R=-\frac{1}{20}{x}^2+25x

\$255

For maximising the revenue 50 units should be sold and the maximum revenue is \$500.

The company should charge \$10 for each product in order to maximise the revenue.

In order to earn at least \$480 as revenue, the company should charge minimum \$8 per product.

Explanation of Solution

Given:

The price p and the quantity x sold of a certain product obey the demand equation

$x=-20p+500,0<p\le 25$

Formula used:

The revenue is the product of the unit selling price and the number of product s sold.

Calculation:

a. We have $x=-20p+500,0<p\le 25$.

Therefore,

$x-500=-20p$

$\Rightarrow p=-\frac{1}{20}x+25$

The revenue,

R=px=\left(-\frac{1}{20}x+25\right)x

$=-\frac{1}{20}{x}^2+25x$

Therefore, R=-\frac{1}{20}{x}^2+25x

b. When $x=20$, we have

R=-\frac{1}{20}{(20)}^2+25(20)=480

When 20 units are sold, the revenue is \$480.

c. The equation of revenue is a quadratic equation with $a=-\frac{1}{20},b=25,c=0$.

Since, $a$ is negative, the vertex is the maximum point of the quadratic function.

Therefore, the maximum point is

$x=-\frac{b}{2a}=-\frac{25}{2(-\frac{1}{20})}=250$

Therefore, for maximising the revenue 250 units must be sold.

The maximum revenue is

$R(-\frac{b}{2a})=R(250)=-\frac{1}{20}{(250)}^2+25(250)=3125$

The maximum revenue by selling the product is \$3125.

d. In order to maximise the revenue, the number of product to be sold is 250.

Therefore, the price of the product should be

$p=-\frac{1}{20}(250)+25=12.5$

Thus, the company should charge $\$12.5$ for each product in order to maximise the revenue.

e. To receive at least \$3000 in revenue, we have

R=-\frac{1}{20}{x}^2+25x\ge 3000

\Rightarrow -\frac{1}{20}{x}^2+25x-3000\ge 0

\Rightarrow x\ge \frac{-25\pm \sqrt{({25}^2)-4(-\frac{1}{20})(-3000)}}{2(-\frac{1}{20})}

$\Rightarrow x\ge \frac{-25\pm 5}{-\frac{1}{10}}$

\Rightarrow x\ge \frac{10(-25\pm 5)}{-1}

$x\ge \frac{10(-25-5)}{-1}$            x\ge \frac{10(-25+5)}{-1}

and

$\Rightarrow x\ge 300$               \Rightarrow x\ge 200

Thus, for getting at least $\$3000$ as revenue, minimum of 300 items or 200 items has to be sold.

Then, the price of each item will be

x=-20p+500\ge 300            x=-20p+500\ge 200

$\Rightarrow -20p\ge 300-500$      and    \Rightarrow -20p\ge 200-500

\Rightarrow p\ge 10                                         \Rightarrow p\ge 15

Therefore, in order to earn at least \$3000 as revenue, the company should charge the product between \$10 and \$15.