Problem 32 - Physics for Scientists and Engineers with Modern Physics

Question

(III) I agree to hire you for 30 days and you can decide between two possible methods of payment: cither \((1) \$ 1000\) a day, or (2) one penny on the first day, two pennics on the scoond day and continue to double your daily pay cach day up to day 30. Use quick estimation to make your decision, and justify it.

Step-by-Step Solution

Answer

Option 2: It's more profitable, totaling approximately \( \$10.7\) million versus \( \$30,000\).

Step 1: Understanding the Payment Options

The problem gives us two payment options. Option 1 pays a flat rate of \( \\(1000 \) per day for 30 days. In Option 2, you're paid \( \\)0.01 \) on the first day, and the amount doubles each subsequent day for 30 days.

Step 2: Calculating the Total Payment for Option 1

For Option 1: Since the payment is the same every day, the calculation is straightforward. The total payment over 30 days is: \( 30 \times 1000 = 30000 \). Therefore, you would earn \( \$30,000 \) in total.

Step 3: Understanding the Growth in Option 2

In Option 2, the payment starts at \( \$0.01 \) or 1 penny on the first day, doubling each day for 30 days. This sequence can be represented by a geometric progression where the first term \( a = 0.01 \) and the common ratio \( r = 2 \).

Step 4: Calculating the Total for Day 30 in Option 2

The payment on any particular day \( n \) is given by the formula: \( a_n = a \cdot r^{n-1} \). On day 30, the payment would be: \( 0.01 \times 2^{29} \).

Step 5: Summing the Series for Option 2

To find the total payment over 30 days for Option 2, sum the geometric series: \( S_n = a \frac{r^n - 1}{r-1} \). For \( n = 30 \), \( a = 0.01 \), and \( r = 2 \), this gives:\[ S_{30} = 0.01 \frac{2^{30} - 1}{2 - 1} = 0.01 \times (2^{30} - 1) \]

Step 6: Approximating the Sum

Calculate \( 2^{30} \), which is quite large; it's approximately \( 1,073,741,824 \). Thus, \( S_{30} \approx 0.01 \times (1,073,741,824 - 1) = 10,737,418.23 \) dollars.

Step 7: Determining the Better Payment Option

Compare the totals from both options. Option 1 brings \( \\(30,000 \), while Option 2 brings substantially more at approximately \( \\)10,737,418.23 \). Thus, Option 2 is clearly more profitable.

Key Concepts

Sequence and Series

When considering Option 2 of the payment plan, the concept of sequences and series is essential. A sequence is a list of numbers that are in a specific, regularly repeating order. For this problem, we identified the pattern as a geometric progression. This type of sequence occurs when each term after the first is found by multiplying the previous one by a constant, known as the common ratio.

Here, the initial amount is a penny, or $0.01. Each day, this amount doubles, making the common factor 2. This leads to a sequence: 0.01, 0.02, 0.04, and so on.

By the 30th day, this doubling pattern causes payments to grow exponentially. To find the total payment over 30 days, we need to sum up all these daily payments. This is where the series concept is used — the sum of the terms in a sequence. By applying the formula for the sum of a geometric series, the total payment is efficiently calculated despite the rapid growth of payments each day.

Exponential Growth

Exponential growth refers to the process of quantities multiplying by a constant rate over consistent intervals. It starts slowly but can quickly escalate, which makes Option 2 a perfect example. Initially, receiving a single penny might seem negligible. However, due to doubling each day, the amount becomes significantly larger very rapidly.

In this particular problem, doubling the pennies day by day means that by day 20, the payment is already a lot more substantial compared to the first day. Exponential functions, such as this doubling sequence, have vast applications in finance, biology, and computing due to their fast-paced growth pattern.

Understanding the criticality of how these numbers rise helps to analyze where exponential growth fits in various real-life contexts. Selecting a payment option that harnesses exponential growth can be advantageous when projected over long periods.

Mathematical Estimation

Being proficient in mathematical estimation allows you to make swift decisions without exact calculations. For this exercise, we used estimation to quickly assess each payment option's practicality.

• Option 1, straightforwardly computed as 30 days multiplied by $1000, results in $30,000.
• Option 2 required understanding of the geometric series. Through a rough estimate of calculating powers of 2, we approximated the result as significantly higher than $30,000.

With larger numbers like 2 to the 30th power, it’s critical to employ estimation to gauge the magnitude. Estimating ensures that decisions can be made quickly when exact values are cumbersome to compute. Practicing this skill can be invaluable, not just academically but in everyday life for budgeting, planning, or even during negotiations.