Q. No. 6 - Precalculus Enhanced with Graphing Utilities

Question

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?

Step-by-Step Solution

Answer

(a) The revenue R as a function of x is $R=-\frac{1}{20}{x}^2+25x$.

(b) The revenue is 480 dollars when 20 units are sold.

(c) The maximum revenue is 3125 dollars when $x=250$.

(d) The company should charge 12.5 dollars per unit to earn maximum revenue.

(e) The company should charge between 10 and 15 dollars per unit to earn at least 3000 dollars in revenue.

Part (a) Step 1. Given information.

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

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

Part (a) Step 2. The revenue as a function of x .

Write p as a function of x.

$$\begin{gathered} x=-20p+500 \\ \frac{x}{-20}=p-25 \\ -\frac{1}{20}x+25=p \\ p=-\frac{1}{20}x+25 \end{gathered}$$

As we know that $R=xp$ then

$$\begin{gathered} R=x(-\frac{1}{20}x+25) \\ =-\frac{1}{20}{x}^2+25x \end{gathered}$$

So the revenue R as a function of x is $R=-\frac{1}{20}{x}^2+25x$.

Part (b) Step 1. Substitute x = 20 in R = - 1 20 x 2 + 25 x .

We get

$$\begin{gathered} R=-\frac{1}{20}{\left(20\right)}^2+25\left(20\right) \\ =-20+500 \\ =480 \end{gathered}$$

So the revenue is 480 dollars when 20 units are sold.

Part (c) Step 1. The maximum revenue.

The function R is a quadratic function with $a=-\frac{1}{20}, b=25$, and $c=0$. Because $a<0$, the vertex is the highest point on the parabola.

The revenue R is maximum when x is

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

Substitute $x=250$ in $R=-\frac{1}{20}{x}^2+25x$.

$$\begin{gathered} R=-\frac{1}{20}{\left(250\right)}^2+25\left(250\right) \\ =-3125+6250 \\ =3125 \end{gathered}$$

So the revenue is maximum at 3125 dollars at $x=250$.

Part (d) Step 1. Substitute x = 250 in p = - 1 20 x + 25 .

We get

$$\begin{gathered} p=-\frac{1}{20}\left(250\right)+25 \\ =-12.5+25 \\ =12.5 \end{gathered}$$

So the company should charge 12.5 dollars to maximize the revenue.

Part (e) Step 1. The price when maximum revenue is 3000 dollars.

Now $R=xp$, where $x=-20p+500$

We get

$$\begin{gathered} R=p(-20p+500) \\ =-20p^2+500p \end{gathered}$$

For revenue to be at least 3000 dollars,

$$\begin{gathered} -20p^2+500p=3000 \\ -p^2+25p=150 \\ {p}^2-25p+150=0 \end{gathered}$$

By quadratic formula,

$$\begin{gathered} p=\frac{-(-25)\pm \sqrt{{(-25)}^2-4\left(150\right)\left(1\right)}}{2\left(1\right)} \\ =\frac{25\pm \sqrt{25}}{2} \\ =\frac{25\pm 5}{2} \end{gathered}$$

We have

$$\begin{gathered} p=\frac{25+5}{2} or p=\frac{25-5}{2} \\ p=15 or p=10 \end{gathered}$$

So the company should charge between 10 and 15 dollars to earn a maximum revenue of 3000 dollars.