Question

3 a wireless company offers two plans for making international calls in a month.

• in plan a they charge a one - time fee of $4 and then charge $0.20 per minute for calls.
• in plan b they charge a one - time fee of $10 and then charge $0.05 per minute for calls.

a. for plan a, write an expression that would give the average cost per minute of making minutes of phone call with that plan during the month.

b. for plan b, write an expression that would give the average cost per minute of making minutes of phone call with that plan during the month.

c. after how many minutes in the month would it be better to switch from plan a to plan b. show all steps taken to solve. if applicable, round to the nearest whole number.

Explanation:

Part (a)

Step1: Determine total cost for Plan A

The total cost for Plan A is the one - time fee plus the cost per minute times the number of minutes. The one - time fee is $4, and the cost per minute is $0.20. If the number of minutes is \(x\), the total cost \(C_A=4 + 0.20x\).

Step2: Find average cost per minute

The average cost per minute is the total cost divided by the number of minutes. So the average cost per minute for Plan A, \(A(x)=\frac{4 + 0.20x}{x}\) (where \(x>0\) since \(x = 0\) would be undefined for average cost calculation).

Step1: Determine total cost for Plan B

The total cost for Plan B is the one - time fee plus the cost per minute times the number of minutes. The one - time fee is $10, and the cost per minute is $0.05. If the number of minutes is \(x\), the total cost \(C_B = 10+0.05x\).

Step2: Find average cost per minute

The average cost per minute is the total cost divided by the number of minutes. So the average cost per minute for Plan B, \(B(x)=\frac{10 + 0.05x}{x}\) (where \(x>0\) since \(x = 0\) would be undefined for average cost calculation).

Step1: Set up the inequality for when Plan B is better

We want to find when the average cost of Plan B is less than the average cost of Plan A. So we set up the inequality \(\frac{4+0.20x}{x}<\frac{10 + 0.05x}{x}\). Since \(x>0\) (number of minutes can't be zero or negative), we can multiply both sides of the inequality by \(x\) (which is positive, so the inequality sign remains the same) to get \(4 + 0.20x<10+0.05x\).

Step2: Solve the inequality

Subtract \(0.05x\) from both sides: \(4+0.20x - 0.05x<10+0.05x-0.05x\), which simplifies to \(4 + 0.15x<10\).
Then subtract 4 from both sides: \(4-4 + 0.15x<10 - 4\), which gives \(0.15x<6\).
Now divide both sides by 0.15: \(x<\frac{6}{0.15}\).
Calculate \(\frac{6}{0.15}=\frac{6}{\frac{15}{100}}=6\times\frac{100}{15}=40\). Wait, no, we made a mistake. Wait, we want to find when Plan B is better, so actually we should find when the total cost of Plan B is less than the total cost of Plan A (maybe the average cost approach was mis - leading, let's re - do it with total cost).

Let's re - define: Let \(C_A=4 + 0.20x\) (total cost for Plan A) and \(C_B=10+0.05x\) (total cost for Plan B). We want to find when \(C_B

So \(10 + 0.05x<4+0.20x\)
Subtract \(0.05x\) from both sides: \(10<4 + 0.15x\)
Subtract 4 from both sides: \(6<0.15x\)
Divide both sides by 0.15: \(x>\frac{6}{0.15}=40\)

Ah, that's the correct approach. Because when we talk about "better" (cheaper), we should compare total costs, not average costs (since average cost has \(x\) in the denominator and can be misleading). So the correct inequality is based on total cost.

So:

Step1: Set total cost of B < total cost of A

\(10 + 0.05x<4+0.20x\)

Step2: Subtract 0.05x from both sides

\(10<4 + 0.15x\)

Step3: Subtract 4 from both sides

\(6<0.15x\)

Step4: Divide by 0.15

\(x > \frac{6}{0.15}=40\)

Answer:

\(\frac{4 + 0.20x}{x}\)

Part (b)