Q. 10 - Precalculus Enhanced with Graphing Utilities

Question

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation $p=-\frac{1}{10}x+1000$ .

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

(b) What is the revenue if 400 units are sold? 

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

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

Step-by-Step Solution

Answer

Part (a). Function is $R\left(x\right)=-\frac{1}{10}{x}^2+1000x$

Part (b). The revenue is 384000

Part (c). 5000 units maximizes revenue and the maximum revenue is 2500000 dollars. 

Part (d). The price is 500 dollars.

Step 1. Given information.

The price p (in dollars) and the quantity x sold of a certain product obey the demand equation $p=-\frac{1}{10}x+1000$.

Part (a) Step 1. Find a model that expresses the revenue R as a function of x.

It is know that the price p (in dollars) and quantity x sold of a certain product is given by equation: 

$p=-\frac{1}{10}x+1000$

Model that express the revenue R(remember: $R=xp$) as a function of x is given by:

$R\left(x\right)=-\frac{1}{10}{x}^2+1000x$

Part (b) Step 1. Find the revenue if 400 units are sold?

The function of the revenue is given by $R\left(x\right)=-\frac{1}{10}{x}^2+1000x$.

$x=400$

$$\begin{gathered} R\left(400\right)=-\frac{1}{10}{\left(400\right)}^2+1000.400 \\ R\left(400\right)=-16000+400000 \\ R\left(400\right)=384000 \end{gathered}$$

The revenue is $384000$ dollars if 400 units are old.

Part (c) Step 1. What quantity x maximizes revenue? What is the maximum revenue?

The function of revenue is given by $R\left(x\right)=-\frac{1}{10}{x}^2+1000x$.

Notice that $a<0$ which means that the function has the maximum value. The maximum value is in the vertex.

The vertex has coordinates:

$\left(-\frac{b}{2a},R\left(-\frac{b}{2a}\right)\right)$

So, $-\frac{b}{2a}=-\frac{1000}{2·\left(-0.1\right)}=5000$

And $R\left(5000\right)=-\frac{1}{10}·{5000}^2+1000·5000=2500000$

Therefore, 5000 units maximizes revenue and the maximum revenue is 2500000 dollars.

Part (d) Step 1. What price should the company charge to maximize revenue?

The revenue function is given by $R\left(x\right)=-\frac{1}{10}{x}^2+1000x$.

Solve $R=xp$if $r=2500000, x=5000$

$$\begin{gathered} R=xp \\ 2500000=5000p \\ p=\frac{2500000}{5000}=500 \end{gathered}$$

Therefore, the price is 500 dollars.