Question

Alicia has invented a new app that two companies are interested in purchasing for a 2-year contract.
Company
\[\text{A}\] is offering
\[\$10{,}000\] for the first month and will increase the amount each following month by
\[\$5000\].
Company
\[\text{B}\] is offering
\[\$500\] for the first month and will double their payment each following month.
For which monthly payment will Company
\[\text{B}\]'s payment first exceed Company
\[\text{A}\]'s payment?

Answer 1

To determine when Company B's payment first exceeds Company A's payment, we need to define the payment structures for both companies over the months.

**Company A:**
- The payment starts at \$10,000 for the first month.
- Each subsequent month, the payment increases by \$5,000.

The payment for Company A in month \( n \) can be expressed as:

\[
A(n) = 10000 + 5000(n - 1)
\]

This simplifies to:

\[
A(n) = 10000 + 5000n - 5000 = 5000n + 5000
\]

**Company B:**
- The payment starts at \$500 for the first month.
- Each subsequent month, the payment doubles.

The payment for Company B in month \( n \) can be expressed as:

\[
B(n) = 500 \times 2^{(n - 1)}
\]

Now, we need to find the smallest \( n \) such that \( B(n) > A(n) \).

Setting up the inequality:

\[
500 \times 2^{(n - 1)} > 5000n + 5000
\]

We can simplify this by dividing everything by 500:

\[
2^{(n - 1)} > 10n + 10
\]

Next, we will calculate values of \( 2^{(n - 1)} \) and \( 10n + 10 \) for several values of \( n \) to find the point where Company B's payment exceeds Company A's.

- For \( n = 1 \):
\[
2^{0} = 1, \quad 10(1) + 10 = 20 \quad \Rightarrow \quad 1 < 20
\]

- For \( n = 2 \):
\[
2^{1} = 2, \quad 10(2) + 10 = 30 \quad \Rightarrow \quad 2 < 30
\]

- For \( n = 3 \):
\[
2^{2} = 4, \quad 10(3) + 10 = 40 \quad \Rightarrow \quad 4 < 40
\]

- For \( n = 4 \):
\[
2^{3} = 8, \quad 10(4) + 10 = 50 \quad \Rightarrow \quad 8 < 50
\]

- For \( n = 5 \):
\[
2^{4} = 16, \quad 10(5) + 10 = 60 \quad \Rightarrow \quad 16 < 60
\]

- For \( n = 6 \):
\[
2^{5} = 32, \quad 10(6) + 10 = 70 \quad \Rightarrow \quad 32 < 70
\]

- For \( n = 7 \):
\[
2^{6} = 64, \quad 10(7) + 10 = 80 \quad \Rightarrow \quad 64 < 80
\]

- For \( n = 8 \):
\[
2^{7} = 128, \quad 10(8) + 10 = 90 \quad \Rightarrow \quad 128 > 90
\]

Thus, Company B's payment first exceeds Company A's payment at \( n = 8 \).

Finally, calculating the actual payments:

- For Company A in month \( 8 \):
\[
A(8) = 5000 \times 8 + 5000 = 40000 + 5000 = 45000
\]

- For Company B in month \( 8 \):
\[
B(8) = 500 \times 2^{7} = 500 \times 128 = 64000
\]

Conclusion: The first month where Company B’s payment exceeds Company A's payment is \( n = 8 \).