You operate a gaming Web site, where users must pay a small fee to log on. When you charged $3 the demand was 520 log-ons per month. When you lowered the price to $2.50, the demand increased to 780 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.
R(x)=
(b) Your Internet provider charges you a monthly fee of $10 to maintain your site. Express your monthly profit P as a function of the log-on fee x.
P(x)=
[What formula do I use to solve this? How should I approach it?]
Determine the log-on fee you should charge to obtain the largest possible monthly profit.
x=$
[Do I plug in some number?]
What is the largest possible revenue?
[How do I find this?]
Thank you.
If we have a linear demand function, it will look like
R = mx+b where x is the price and y is the demand at that price.
780 = 2.5m + b
520 = 3m + b
so,
-260 = .5m
m = -520
b = 2080
R = 2080 - 520x
profit = revenue - cost
revenue = demand * price
P = R*x - 10
P = 2080x - 520x^2 - 10
you have a parabola, where the vertex is at x =
-b/2a = 2080/1040 = 2
P(2) = 2070
To solve this problem, you can use linear regression to find the demand equation and then use that to calculate the revenue and profit functions.
(a) To construct a linear demand function, we can use the formula for a line, y = mx + b, where y represents the demand (log-ons per month), x represents the log-on fee, m represents the slope, and b represents the y-intercept.
Let's start by finding the slope (m) using the given data points (x₁, y₁) = ($3, 520) and (x₂, y₂) = ($2.50, 780):
m = (y₂ - y₁) / (x₂ - x₁)
= (780 - 520) / ($2.50 - $3)
= 260 / $0.50
= 520
Now, we can find the y-intercept (b) by substituting one of the data points into the equation:
520 = 520($3) + b
520 = 1560 + b
b = 520 - 1560
b = -1040
Therefore, the demand equation for your Web site is:
D(x) = 520x - 1040
To obtain the monthly revenue (R) as a function of the log-on fee (x), we can multiply the demand equation by the log-on fee:
R(x) = x * D(x)
= x * (520x - 1040)
= 520x² - 1040x
(b) To express the monthly profit (P) as a function of the log-on fee (x), we subtract the monthly fee ($10) from the revenue:
P(x) = R(x) - $10
= 520x² - 1040x - $10
To determine the log-on fee that results in the largest possible monthly profit, we need to find the maximum value of the profit function (P(x)). One way to do this is by finding the vertex of the quadratic equation.
The x-coordinate of the vertex (x-value for maximum profit) can be found using the formula:
x = -b / (2a)
For the quadratic equation P(x) = 520x² - 1040x - $10, a = 520 and b = -1040. Substituting these values into the formula, we have:
x = -(-1040) / (2 * 520)
= 1040 / 1040
= 1
Therefore, the log-on fee you should charge to obtain the largest possible monthly profit is $1.
To find the largest possible revenue, we can substitute the log-on fee ($1) into the revenue equation:
R(x) = 520x² - 1040x
R($1) = 520($1)² - 1040($1)
= 520 - 1040
= -$520
The largest possible revenue is -$520, which implies that the maximum revenue is negative, indicating that the log-on fee is too low to cover the costs and generate a profit.
It's important to note that the given data only provides a linear approximation of the demand function. Based on this approximation, the log-on fee of $1 results in the largest possible profit. However, analyzing the data further and considering other factors might be necessary to make a more accurate decision.
To solve this problem, we need to use the concept of demand functions and revenue functions. The demand function represents the relationship between the price and the quantity demanded, while the revenue function calculates the total revenue generated at a given price.
(a) To construct a linear demand function, we can use the formula for the equation of a straight line: y = mx + b, where m is the slope and b is the y-intercept.
We are given two data points: (3, 520) and (2.50, 780). Using these points, we can find the slope:
m = (780 - 520) / (2.50 - 3)
= 260 / (-0.50)
= -520
Now that we have the slope, we can substitute one of the points into the equation and find the y-intercept. Let's use the point (3, 520):
520 = -520(3) + b
520 = -1560 + b
b = 2080
Therefore, the linear demand function for your website is:
D(x) = -520x + 2080
To find the monthly revenue R as a function of the log-on fee x, we multiply the demand function by x because revenue is the product of price and quantity demanded:
R(x) = x * D(x)
R(x) = x*(-520x + 2080)
R(x) = -520x^2 + 2080x
(b) The monthly profit P is calculated by subtracting the cost (monthly fee) from the monthly revenue:
P(x) = R(x) - cost = -520x^2 + 2080x - 10
To determine the log-on fee that maximizes monthly profit, we need to find the maximum point on the profit function. We can use calculus to find the maximum point by taking the derivative of the profit function and setting it equal to zero:
P'(x) = -1040x + 2080 = 0
Solving for x, we get:
x = 2
Therefore, the log-on fee you should charge to obtain the largest possible monthly profit is $2.
To find the largest possible revenue, we substitute the log-on fee x = 2 into the revenue function we obtained in part (a):
R(2) = -520(2)^2 + 2080(2)
R(2) = -2080 + 4160
R(2) = $2080
Hence, the largest possible revenue is $2080.
Please note that this analysis assumes a linear demand function and may not account for other factors affecting demand or pricing strategies you may want to consider.