Chapter 3, Problem 10CT

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?

To determine

To calculate:

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 $x$ maximises the revenue? What is the maximum revenue?

d. What price should the company charge to maximise the revenue.

Answer to Problem 10CT

Solution:

a. $R=-\frac{1}{10}{x}^2+1000x$

b. $\$384000$

c. The maximum revenue $\$2500000$ is obtained by selling 5000 products.

d. $\$500$

Explanation of Solution

Given:

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

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

Formula used:

He revenue is the product of the number items sold and the price per unit item.

Consider the function $f(x)=a{x}^2+bx+c,a\ne 0$.

The vertex is at $\left(\frac{-b}{2a},f\left(\frac{-b}{2a}\right)\right)$

If $a>0$, then the graph opens up and the vertex is the minimum point.

If $a<0$, then the graph opens downwards and the vertex is the maximum point.

Calculation:

a. The revenue is the product of $x$ and $p$. Thus, we get

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

Thus, the revenue equation is $R=-\frac{1}{10}{x}^2+1000x$

b. When $x=400$, we get

$R=-\frac{1}{10}{(400)}^2+1000(400)=384000$

Thus, the revenue will be $\$384000$ if 400 units are sold.

c. We can see that the revenue equation is a quadratic function with $a=-0.1,b=1000,c=0$

Since, $a$ is negative, the function is having its maximum at its vertex.

Thus, the maximum value is at

$x=\frac{-b}{2a}=\frac{-1000}{2(-0.1)}=5000$

The maximum value is

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

Therefore, the maximum revenue $\$2500000$ is obtained by selling 5000 products.

d. To maximise the revenue, the company has to sell 5000 products. Then the price of each item should be

$p(5000)=-\frac{1}{10}(5000)+1000=500$

Thus, for maximising the revenue, the price of each product should be $\$500$.