Part (a): Constructing the Linear Demand Function and Revenue Function

You are given two points on the demand function, where $x$ is the price per log-on and $d(x)$ is the number of log-ons:

• When the price was $3, the demand was 510 log-ons per month, i.e., $(x_1, d_1) = (3, 510)$.
• When the price was $2.50, the demand was 765 log-ons per month, i.e., $(x_2, d_2) = (2.50, 765)$.

Since the demand function is linear, it takes the form $d(x) = mx + b$, where $m$ is the slope and $b$ is the intercept.

Step 1: Calculate the slope $m$

The slope $m$ of the linear demand function is given by: $m = \frac{d_2 - d_1}{x_2 - x_1} = \frac{765 - 510}{2.50 - 3} = \frac{255}{-0.5} = -510$

Step 2: Write the equation of the demand function

Using the point-slope form of the line, $d(x) = m x + b$, we now substitute $m = -510$ and one of the points (for example, $(3, 510)$) to solve for $b$.

$510 = -510(3) + b$ $510 = -1530 + b$ $b = 2040$

Thus, the demand function is: $d(x) = -510x + 2040$

Step 3: Construct the revenue function $R(x)$

Revenue $R(x)$ is the product of the price per log-on $x$ and the demand $d(x)$, i.e., $R(x) = x \cdot d(x)$ Substitute $d(x) = -510x + 2040$ into the revenue formula: $R(x) = x(-510x + 2040)$ $R(x) = -510x^2 + 2040x$

Part (b): Constructing the Profit Function

Your monthly profit $P(x)$ is the difference between the revenue and the cost. The only cost is the $40 monthly fee charged by your Internet provider. Thus, the profit function is:

$P(x) = R(x) - 40$ Substitute $R(x) = -510x^2 + 2040x$ into the profit formula: $P(x) = (-510x^2 + 2040x) - 40$ $P(x) = -510x^2 + 2040x - 40$

Step 3: Maximizing the Profit

To find the price $x$ that maximizes the profit, we take the derivative of $P(x)$ and set it equal to zero.

$P'(x) = \frac{d}{dx}(-510x^2 + 2040x - 40) = -1020x + 2040$ Set $P'(x) = 0$ to find the critical point: $-1020x + 2040 = 0$ $1020x = 2040$ $x = \frac{2040}{1020} = 2$

Thus, the price that maximizes the profit is $2.

Step 4: Calculate the Maximum Profit

Substitute $x = 2$ into the profit function $P(x) = -510x^2 + 2040x - 40$ to find the maximum profit:

$P(2) = -510(2)^2 + 2040(2) - 40$ $P(2) = -510(4) + 4080 - 40$ $P(2) = -2040 + 4080 - 40 = 2000$

Thus, the largest possible monthly profit is $2000.

Summary of Answers:

(a) The monthly revenue as a function of the log-on fee is: $R(x) = -510x^2 + 2040x$

(b) The monthly profit as a function of the log-on fee is: $P(x) = -510x^2 + 2040x - 40$

• The log-on fee to maximize profit is $x = 2$ dollars.
• The largest possible monthly profit is $2000.

---

Would you like more details on any step or further clarifications?

Related Questions:

1. What happens to the profit if the log-on fee is set to $2.50 instead of $2?
2. How would the demand function change if new data indicated different log-on demand at higher or lower prices?
3. If the fixed monthly cost increases, how will the profit-maximizing price change?
4. What is the break-even price, i.e., the price at which profit is zero?
5. How would a shift in customer behavior (e.g., increased demand elasticity) affect the demand function?

Tip:

When optimizing revenue or profit functions, always look for ways to derive the maximum by taking derivatives and solving for critical points.