Chapter 3.4, Problem 5AYU

Maximizing Revenue The price $p$ (in dollars) and the quantity $x$ sold of a certain product obey the demand equation

$x=-5p+100$, $0<p\le 20$

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

(b) What is the revenue if 15 units are sold?

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

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

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

To determine

To calculate:

The revenue as a function of x.

What is the revenue if 15 units are sold?

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

What price should the company charge to maximise the revenue?

What price should the company charge to earn at least $480 in revenue?

Answer to Problem 5AYU

Solution:

R=-\frac{1}{5}{x}^2+20x

\$255

For maximising the revenue 50 units should be sold and the maximum revenue is \$500.

The company should charge \$10 for each product in order to maximise the revenue.

In order to earn at least \$480 as revenue, the company should charge minimum \$8 per product.

Explanation of Solution

Given:

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

$x=-5p+100,0<p\le 20$

Formula used:

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

Calculation:

a. We have $x=-5p+100,0<p\le 20$.

Therefore,

$x-100=-5p$

$\Rightarrow p=-\frac{1}{5}x+20$

The revenue,

$R=px=\left(-\frac{1}{5}x+20\right)x$

$=-\frac{1}{5}{x}^2+20x$

Therefore $R=-\frac{1}{5}{x}^2+20x$

b. When $x=15$, we have

R=-\frac{1}{5}{(15)}^2+20(15)=255

When 15 units are sold, the revenue is \$255.

c. The equation of revenue is a quadratic equation with $a=-\frac{1}{5},b=20,c=0$.

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

Therefore, the maximum point is

$x=-\frac{b}{2a}=-\frac{20}{2(-\frac{1}{5})}=50$

Therefore, for maximising the revenue 50 units must be sold.

The maximum revenue is

$R(-\frac{b}{2a})=R(50)=-\frac{1}{5}{(50)}^2+20(50)=500$

The maximum revenue by selling the product is $\$500$.

d. In order to maximise the revenue, the number of product to be sold is 50.

Therefore, the price of the product should be

$p=-\frac{1}{5}(50)+20=10$

Thus, the company should charge $\$10$ for each product in order to maximise the revenue.

e. To receive at least $\$480$ in revenue, we have

$-\frac{1}{5}{x}^2+20x\ge 480$

\Rightarrow -\frac{1}{5}{x}^2+20x-480\ge 0

\Rightarrow x=\frac{-20\pm \sqrt{({20}^2)-4(-\frac{1}{5})(-480)}}{2(-\frac{1}{5})}

\Rightarrow x\ge \frac{-20\pm 4}{-\frac{2}{5}}

\Rightarrow x\ge \frac{5(-20\pm 4)}{-2}

x\ge \frac{5(-20-4)}{-2}          x\ge \frac{5(-20+4)}{-2}

and

\Rightarrow x\ge 60               \Rightarrow x\ge 40

Thus, for getting at least $\$480$ as revenue, minimum of 60 items or 40 items has to be sold.

Then, the price of each item will be

x=-5p+100\ge 60           x=-5p+100\ge 40

\Rightarrow -5p\ge 60-100     and    \Rightarrow -5p\ge 40-100

\Rightarrow p\ge 8                                      \Rightarrow p\ge 12

Therefore, in order to earn at least \$480 as revenue, the company should charge the product between \$8 and \$12.