Chapter 3, Problem 23RE

23. Demand Equation the price p (in dollars) and the quantity x sold of a certain product obey the demand equation

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

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

(b) What is the revenue if 100 units are sold?

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

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

To determine

To calculate:

a. The function that relates the revenue $R$ as a function of $x$.

b. What is the revenue if 100 units are sold?

c. What quantity $x$ maximises the revenue and what is the maximum revenue?

d. What price should the company charge to maximise the revenue?

Answer to Problem 23RE

Solution:

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

b. If 100 units are sold, the revenue will be $\$14000$.

c. The maximum revenue is $\$56250$ and is obtained when 750 units are sold.

d. In order for the company to maximise the revenue, the price should be $\$75$.

Explanation of Solution

Given:

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

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

Formula used:

The revenue is the product of the quantity of units sold and the price of each individual unit.

For a quadratic function $f(x)=a{x}^2+bx+c$, the vertex $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$ is the maximum point if $a$ is negative and the vertex is the minimum point if $a$ is positive.

Calculation:

a. The revenue function is

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

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

b. When $x=100$, we get

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

$=14000$

c. Here, from the revenue function, we can see that $a=-\frac{1}{10}<0$ is negative, thus, the vertex is its maximum point.

Thus, we have

$\frac{-b}{2a}=\frac{-150}{2(-0.1)}=750$

$R\left(\frac{-b}{2a}\right)=R(750)=-\frac{1}{10}{(750)}^2+150(750)=56250$

Therefore, the maximum revenue is $\$56250$ and is obtained when 750 units are sold.

d. In order to maximise the revenue, the price of each unit should be

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

$=75$