Chapter 3.4, Problem 4AYU

Maximizing Revenue The price $p$ (in dollars) and the quantity $A$ 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?

To determine

To calculate:

The model that expresses the revenue R as a function of x.

What is the domain of $\text{R}$?

What is the revenue if 100 units are sold?

What quantity of x maximises the revenue and what is the maximum revenue.

What price should the company charge to maximise the revenue.

Answer to Problem 4AYU

Solution:

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

$0\le x\le 300$

\$9,999.3333

For maximising the revenue 150 units should be sold and the maximum revenue is \$7500.

The company should charge \$50 for each product in order to maximise the revenue.

Explanation of Solution

Given:

The price p and the quantity x sold of a certain product obey the demand equation

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

Formula used:

The revenue is the product of the unit selling price and the number of product s sold.

Calculation:

a. The revenue,

R=px=\left(-\frac{1}{3}x+100\right)x

$=-\frac{1}{3}{x}^2+100x$

Therefore, $R=-\frac{1}{3}{x}^2+100x$

b. We know that the number of the products sold will never be negative. Therefore, we have $x\ge 0$. And the price of the product will also never be negative, thus, we get $p\ge 0$. Thus,

$p\ge 0$

$\Rightarrow -\frac{1}{3}x+100\ge 0$

$\Rightarrow 100\ge \frac{1}{3}x$

$\Rightarrow 300\ge x$

Combining both the inequalities of x, we get the domain of R to be $0\le x\le 300$.

c. When $x=100$, we get

R=-\frac{1}{3}(100{)}^2+100(100)

=9,999.3333

Thus, if 200 units are sold, the n the revenue will be \$9,999.3333.

d. The equation of revenue is a quadratic equation with a=-\frac{1}{3},b=100,c=0.

Since, $a$ is negative, the vertex is the maximum point of the quadratic function.

Therefore, the maximum point is

$x=-\frac{b}{2a}=-\frac{100}{2(-\frac{1}{3})}=150$

Therefore, for maximum revenue 150 units should be sold.

The maximum revenue is

$R(-\frac{b}{2a})=R(150)=R=-\frac{1}{3}{(150)}^2+100(150)=7500$

The maximum revenue by selling the product is \$7500.

e. In order to maximise the revenue, the number of product to be sold is 150.

Therefore, the price of the product should be

$p=-\frac{1}{3}(150)+100=50$

Thus, the company should charge \$50 for each product in order to maximise the revenue.