To solve this problem, let's tackle it step-by-step.

(a) Constructing a Linear Demand Function and Monthly Revenue $R$

We are given two price-demand points:

1. At $x = 4$, the demand is 590 log-ons per month.
2. At $x = 3.50$, the demand is 885 log-ons per month.

Assuming a linear demand function of the form: $d(x) = mx + b$ where $d(x)$ is the demand as a function of the log-on fee $x$.

Step 1: Find the slope $m$

Using the two points $(4, 590)$ and $(3.5, 885)$: $m = \frac{885 - 590}{3.5 - 4} = \frac{295}{-0.5} = -590$

Step 2: Find the $y$-intercept $b$

Substitute one of the points, say $(4, 590)$, into the equation $d(x) = -590x + b$: $590 = -590(4) + b$ $590 = -2360 + b$ $b = 2950$

Thus, the demand function is: $d(x) = -590x + 2950$

Step 3: Find the Revenue Function $R(x)$

Revenue is the product of the price $x$ and demand $d(x)$: $R(x) = x \cdot d(x) = x \cdot (-590x + 2950)$ $R(x) = -590x^2 + 2950x$

So, the monthly revenue function is: $R(x) = -590x^2 + 2950x$

(b) Constructing the Profit Function $P(x)$

Your monthly costs are $40. Profit is revenue minus costs, so: $P(x) = R(x) - 40$ $P(x) = (-590x^2 + 2950x) - 40$ $P(x) = -590x^2 + 2950x - 40$

Thus, the profit function is: $P(x) = -590x^2 + 2950x - 40$

(c) Maximizing Profit

To find the log-on fee $x$ that maximizes profit, we need to find the vertex of the parabola represented by $P(x) = -590x^2 + 2950x - 40$. The maximum value of a quadratic function $ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.

Step 1: Apply the vertex formula

Here, $a = -590$ and $b = 2950$: $x = -\frac{2950}{2 \cdot -590} = \frac{2950}{1180} = 2.5$

So, the log-on fee $x$ that maximizes profit is: $x = 2.5$

Step 2: Calculate the Maximum Profit

Substitute $x = 2.5$ into $P(x)$: $P(2.5) = -590(2.5)^2 + 2950(2.5) - 40$ $P(2.5) = -590(6.25) + 2950(2.5) - 40$ $P(2.5) = -3687.5 + 7375 - 40$ $P(2.5) = 3647.5$

Final Answers

1. The revenue function $R(x)$ is: $R(x) = -590x^2 + 2950x$

2. The profit function $P(x)$ is: $P(x) = -590x^2 + 2950x - 40$

3. The log-on fee that maximizes profit is: $x = 2.5$

4. The largest possible monthly profit is: $P(2.5) = 3647.5$

Would you like further details on any step, or help with another topic?

Here are some related questions:

1. How does changing the monthly fee affect profit maximization?
2. What if the demand function was not linear? How would that impact revenue and profit functions?
3. How would an increase in the Internet provider's fee change the optimal log-on fee?
4. Can you determine the price elasticity of demand based on the demand function?
5. How would a decrease in log-on price below $2.5 affect the business?

Tip: Always double-check calculations for profit maximization, as small errors can affect the optimal solution.