Q. No. 4 - 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

$p=-\frac{1}{3}x+100$

(a) Find a model that expresses the revenue R as a function of x.

(b) What is the domain of R?

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

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

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

Step-by-Step Solution

Answer

(a) The revenue R as a function of x is $R=-\frac{1}{3}{x}^2+100x$.

(b) The domain of R is $[0,300)$.

(c) The revenue is 6666.66 dollars when 100 units are sold.

(d) The maximum revenue is 75,000 dollars when $x=150$.

(e) The company should charge 50 dollars to maximize the revenue.

Part (a) Step 1. Given information.

Given a equation for the  price p (in dollars) and the quantity x sold of a certain product as $p=-\frac{1}{3}x+100$.

Part (a) Step 2. A model to express the revenue as a function of x .

As $R=px$ which means

$$\begin{gathered} R=x(-\frac{1}{3}x+100) \\ =-\frac{1}{3}{x}^2+100x \end{gathered}$$

The revenue R as a function of x is $R=-\frac{1}{3}{x}^2+100x$.

Part (b) Step 1. The domain of R .

As x represents the number of quantity sold, we have price $p>0$, so $-\frac{1}{3}x+100>0$.

Solving this linear inequality, we find that 

$$\begin{gathered} -\frac{1}{3}x+100>0 \\ 100>\frac{1}{3}x \\ 300>x \\ x<300 \end{gathered}$$

In addition, quantity sold $x\ge 0$.

Combining these inequalities, we have the domain of R is $\{x:0\le x<300\}=[0,300)$.

Part (c) Step 1. The revenue when 100 units are sold.

Substitute $x=100$ in $R=-\frac{1}{3}{x}^2+100$.

$$\begin{gathered} R=-\frac{1}{3}{\left(100\right)}^2+100\left(100\right) \\ =-\frac{10000}{3}+10000 \\ =\frac{-10000+30000}{3} \\ =\frac{20000}{3} \\ =6666.66 \end{gathered}$$

The revenue is 6666.66 dollars when 100 units are sold.

Part (d) Step 1. The maximum revenue.

The function R is a quadratic function with $a=-\frac{1}{3}, b=100$, and $c=0$. Because $a<0$, the vertex is the highest point on the parabola.

The revenue R is maximum when x is

$$\begin{gathered} x=-\frac{b}{2a} \\ =\frac{-100}{2(-{\displaystyle \frac{1}{3}})} \\ =50\left(3\right) \\ =150 \end{gathered}$$

The revenue maximizes when 150 units are sold.

The maximum revenue is

$$\begin{gathered} R=-\frac{1}{3}{x}^2+100x \\ =-\frac{1}{3}{\left(150\right)}^2+100\left(150\right) \\ =-7500+15000 \\ =75,000 \end{gathered}$$

The maximum revenue is 75,000 dollars.

Part (e) Step 1. The price to maximize the revenue.

The revenue maximizes when 150 units are sold.

So to maximize the price, substitute $x=150$ in $p=-\frac{1}{3}x+100$.

$$\begin{gathered} p=-\frac{1}{3}\left(150\right)+100 \\ =-50+100 \\ =50 \end{gathered}$$

So the company should charge 50 dollars to maximize the revenue.