Q. No. 3 - 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:

$p=-\frac{1}{6}x+100$

(a) Find a model that expresses the revenue R as a function of x (Remember, $R=xp$.)

(b) What is the domain of R?

(c) What is the revenue if 200 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 $R=-\frac{1}{6}{x}^2+100x$.

(b) The domain of R is $[0,600)$.

(c) The revenue on selling 200 units is 13333 dollars (approximately).

(d) The revenue maximizes at $x=300$ and $R\left(300\right)=15000$.

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

Part (a) Step 1. Given information.

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

$p=-\frac{1}{6}x+100$

Part (a) Step 2. Find the equation for revenue using R = x p

We get

$$\begin{gathered} R=x(-\frac{1}{6}x+100) \\ R=-\frac{1}{6}{x}^2+100x \end{gathered}$$

The revenue R as a function of x is $R=-\frac{1}{6}{x}^2+100x$.

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

The domain will be the value of x which R takes.

As x represents the number of quantities sold then $x\ge 0$.

Also the price of quantity is always non-negative then $p>0$.

We get $-\frac{1}{6}x+100>0$.

On solving,

$$\begin{gathered} -\frac{1}{6}x+100>0 \\ 100>\frac{1}{6}x \\ 6\left(100\right)>x \\ 600>x \\ x<600 \end{gathered}$$

So the domain of revenue R is $\left\{x:0\le x<600\right\}=[0,600)$.

Part (c) Step 1. Substitute x = 200 in R = - 1 6 x 2 + 100 x .

We have

$R\left(x\right)=-\frac{1}{6}{x}^2+100x$ then

$$\begin{gathered} R\left(200\right)=-\frac{1}{6}{\left(200\right)}^2+100\left(200\right) \\ =-\frac{20000}{3}+20000 \\ =\frac{-20000+60000}{3} \\ =\frac{40000}{3} \\ =13333.333 \\ \approx 13333 \end{gathered}$$

The revenue on selling 200 units is 13333 dollars (approximately).

Part (d) Step 1. Find the value of x at which revenue maximizes and maximum revenue.

The function R is a quadratic function with $a=-\frac{1}{6}, b=100$ and $c=0$.

As $a<0$, the vertex is the highest point on the parabola.

The revenue R is maximum when the quantity x is

$$\begin{gathered} x=-\frac{b}{2a} \\ =\frac{-100}{2(-{\displaystyle \frac{1}{6}})} \\ =\frac{100}{{\displaystyle \frac{1}{3}}} \\ =100\left(3\right) \\ =300 \end{gathered}$$

and

$$\begin{gathered} R\left(300\right)=-\frac{1}{6}{\left(300\right)}^2+100\left(300\right) \\ =-15000+30000 \\ =15000 \end{gathered}$$

So, the revenue maximizes at $x=300$ and maximum revenue is $15000$ dollars.

Part (e) Step 1. Find the value of p which maximizes revenue.

As calculated revenue maximizes at $x=300$.

Substitute $x=300$ in $p=-\frac{1}{6}x+100$.

We get

$$\begin{gathered} p=-\frac{1}{6}\left(300\right)+100 \\ =-50+100 \\ =50 \end{gathered}$$

The company should charge 50 dollars to maximize the revenue.