Question

You operate a gaming Web site, www.mudbeast.net, where users must pay a small fee to log on. When you charged $3 the demand was 1100 log-ons per month. When you lowered the price to $2.50, the demand increased to 1375 log-ons per month.
A) Construct a linear demand function for your Web site and hence obtain the monthly revenue R as a function of the log-on fee x.
B) Your Internet provider charges you a monthly fee of $30 to maintain your site. Express your monthly profit P as a function of the log-on fee x.
C) Determine the log-on fee you should charge to obtain the largest possible monthly profit. What is the largest possible monthly profit?

Answer 1

Answer:

A) The linear relation between price and demand is:

$d=-550x+2750$

The revenue R is:

$R=-550x^2+2750x$

B) The profit functionP is:

$P=-550x^2+2750x-30$

C) The largest monthly profit is obtained with a log-on fee of $2.5 per month. This corresponds to a profit of $3407.5.

Step-by-step explanation:

We have a site where the number of log-ons depends on our monthly fee. A linear relation is established between the price (log-on fee) and the number of log-ons.

We have two points for this linear relationship:

• At price x=3, the demand is d=1100.
• At price x=2.5, the demand is d=1375.

We will model the relation:

$d=mx+b$

We can calculate the slope m as:

$m=\dfrac{\Delta d}{\Delta x}=\dfrac{d_2-d_1}{x_2-x_1}=\dfrac{1375-1100}{2.5-3}\\\\\\m=\dfrac{275}{-0.5}=-550$

Then, replacing one point in the linear equation, we can calculate the intercept b:

$d_1=mx_1+b\\\\1100=(-550)\cdot 3+b\\\\1100=-1650+b\\\\b=1100+1650=2750$

Then, the linear relation between demand and price is:

$d=-550x+2750$

The revenue R can be expressed as the multiplication of the price and the demand:

$R=x\cdot d=x(-550x+2750)=-550x^2+2750x$

If we have a fixed cost of $30 per month, the profit P is:

$P=R-FC=-550x^2+2750x-30$

We can maximize the profit by deriving the profit function and making it equal to zero.

$\dfrac{dP}{dx}=0\\\\\\\dfrac{dP}{dx}=-550(2x)+2750(1)=0\\\\\\-1100x+2750=0\\\\x=\dfrac{2750}{1100}=2.5$

This corresponds to a profit of:

$P(2.5)=-550(2.5)^2+2750(2.5)-30\\\\P(2.5)=-550\cdot 6.25+6875-30\\\\P(2.5)=-3437.5+6875-30\\\\P(2.5)=3407.5$