Q. No. 5 - 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

$x=-5p + 100; 0<p\le 20$

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

(b) What is the revenue if 15 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 $480 in revenue?

Step-by-Step Solution

Answer

(a) The revenue R as a function of x is $R=-\frac{1}{5}{x}^2+20x$.

(b) The revenue is 255 dollars when 15 units are sold.

(c) The maximum revenue is 500 dollars when $x=50$.

(d) The company should charge 10 dollars per unit to maximize the revenue.

(e) The company should charge between 8 dollars and 12 dollars to earn at least 480 dollars in revenue.

Part (a) Step 1. Given information.

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

$x=-5p+100; 0<p\le 20$

Part (a) Step 2. Express p as a function of x .

We get

$$\begin{gathered} x=-5p+100 \\ \frac{x}{-5}=p+\frac{100}{-5} \\ -\frac{1}{5}x=p-20 \\ -\frac{1}{5}x+20=p \\ p=-\frac{1}{5}x+20 \end{gathered}$$

Part (a) Step 3. Write revenue as a function of x .

As we know that $R=px$ then

$$\begin{gathered} R=x(-\frac{1}{5}x+20) \\ =-\frac{1}{5}{x}^2+20x \end{gathered}$$

Part (b) Step 1. Substitute x = 15 in R = - 1 5 x 2 + 20 x .

We get

$$\begin{gathered} R=-\frac{1}{5}{\left(15\right)}^2+20\left(15\right) \\ =-45+300 \\ =255 \end{gathered}$$

So the revenue is 255 dollars when 15 units are sold.

Part (c) Step 1. The maximum revenue.

The function R is a quadratic function with $a=-\frac{1}{5}, b=20$, and $c=0$. Because $a<0$, the vertex is the highest point on the parabola.

The revenue R is a maximum when x is

$$\begin{gathered} x=-\frac{b}{2a} \\ =-\frac{20}{2(-{\displaystyle \frac{1}{5}})} \\ =50 \end{gathered}$$

The maximum revenue is

$$\begin{gathered} R=-\frac{1}{5}{\left(50\right)}^2+20\left(50\right) \\ =-500+1000 \\ =500 \end{gathered}$$

$R=500$ dollars is maximum.

Part (d) Step 1. Substitute x = 50 in p = - 1 5 x + 20 .

We get

$$\begin{gathered} p=-\frac{1}{5}\left(50\right)+20 \\ =-10+20 \\ =10 \end{gathered}$$

So the company should charge 10 dollars per unit to maximize revenue.

Part (e) Step 1. The charge per unit to earn at least 480 dollars in revenue.

As $R=xp$ then

$$\begin{gathered} R=p(-5p+100) \\ =-5p^2+100p \end{gathered}$$

Now for revenue at least 480 dollars,

$$\begin{gathered} -5p^2+100p=480 \\ -p^2+20p=96 \\ -p^2+20p-96=0 \\ {p}^2-20p+96=0 \end{gathered}$$

By quadratic formula,

$$\begin{gathered} p=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{-(-20)\pm \sqrt{{(-20)}^2-4\left(1\right)\left(96\right)}}{2\left(1\right)} \\ =\frac{20\pm \sqrt{400-384}}{2} \\ =\frac{20\pm 4}{2} \end{gathered}$$

we have

$$\begin{gathered} p=\frac{20+4}{2} or p=\frac{20-4}{2} \\ p=12 or p=8 \end{gathered}$$

So the company should charge between 8 and 12 dollars to earn at least 480 dollars revenue.