Q. 23 - Precalculus Enhanced with Graphing Utilities

Question

Demand Equation The price p (in dollars) and the quantity

x sold of a certain product obey the demand equation

$p=-\frac{1}{10}x+150 0\le x\le 1500$

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

(b) What is the revenue if 100 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

(a) $R=-\frac{1}{10}{x}^2+150x$

(b) $R=14000$

(c) $x=750 \& R=56250$

(d) To maximize revenue, price  $=\$75$

Part (a) Step 1. Given Information

$p=-\frac{1}{10}x+150$

Part (a) Step 2. Calculation

$$\begin{gathered} R=xp \\ =x(-\frac{1}{10}x+150) \\ =(-\frac{1}{10}{x}^2+150x) \end{gathered}$$

Part (b) Step 1. Given Information

$p=-\frac{1}{10}x+150$

Part (b) Step 2. Calculation

For $x=100,$

$R=-\frac{1}{10}\times 10000+150\times 100$

$=14000$

Part (c) Step 1. Given Information

$p=-\frac{1}{10}x+150$

Part (c) Step 2. Calculation

To maximize revenue,

$R\text{'}\left(x\right)=-\frac{1}{5}x+150$, which is equal to zero

$$\begin{gathered} or,-\frac{1}{5}x+150=0 \\ or,x=750 \end{gathered}$$

and maximum revenue is $R=56250$

Part (d) Step 1. Calculation

$p=-\frac{1}{10}x+150$

Part (d) Step 2. Calculation

To maximize revenue $x=750$

Then $p=-\frac{1}{10}x+150$

or, $$\begin{gathered} p=-\frac{1}{10}\times 750+150 \\ =75 \end{gathered}$$