Problem 59
Question
Growth of an Exponential Function Suppose you are offered a job that lasts one month, and you are to be very well paid. Which of the following methods of payment is more profitable for you? (a) One million dollars at the end of the month (b) Two cents on the first day of the month, 4 cents on the second day, 8 cents on the third day, and, in general, \(2^{n}\) cents on the \(n\) th day
Step-by-Step Solution
Verified Answer
Option (b) is more profitable, yielding $21,474,836.46.
1Step 1: Understand the Payment Methods
We are given two options for payment: (a) a fixed amount of one million dollars at the end of the month, and (b) an exponentially growing payment starting at 2 cents on the first day and doubling every day for 30 days.
2Step 2: Analyze Option (a)
In option (a), you will receive a total of one million dollars or $1,000,000 at the end of the month.
3Step 3: Calculate the Total for Option (b)
For option (b), the payment on the nth day is given by the exponential function as \(2^n\) cents. We need to calculate the total amount of cents received over 30 days, which is the sum: \(2^1 + 2^2 + 2^3 + \.\.\. + 2^{30}\).
4Step 4: Use the Formula for the Sum of a Geometric Series
The sum of this geometric series can be calculated using the formula \( S = a \cdot (r^n - 1)/(r-1) \), where \(a\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms. Here, \( a = 2 \), \( r = 2 \), and \( n = 30 \).
5Step 5: Perform the Calculation
Substitute the values into the formula: \( S = 2 \cdot (2^{30} - 1)/(2 - 1) = 2^{31} - 2 \). This result must be divided by 100 to convert cents to dollars.
6Step 6: Compare the Two Amounts
Calculate \(2^{31} - 2 = 2,147,483,646\) cents which is \$21,474,836.46. Compare this result with the one million dollars from option (a).
7Step 7: Conclusion
Since $21,474,836.46 from option (b) is considerably more than $1,000,000 from option (a), option (b) is more profitable.
Key Concepts
Geometric SeriesExponential FunctionPayment Calculation
Geometric Series
Imagine starting with a tiny payment that grows larger every day. This is the essence of a geometric series. A geometric series is a sequence in which each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In our exercise, the payment series starts at 2 cents on the first day, and it increases by a factor of 2 each subsequent day; thus, the common ratio is 2.
To find the total sum of such a series over 30 days, we use the geometric series formula:
To find the total sum of such a series over 30 days, we use the geometric series formula:
- First term, \(a = 2\) cents.
- Common ratio, \(r = 2\).
- Number of terms, \(n = 30\).
Exponential Function
Exponential functions are mathematical expressions that show rapid increases (or decreases). In our problem, the payment for option (b) is described by the exponential function \(2^n\), where "\(n\)" represents the day of the month. This means each day, the payment amount doubles the previous day's payment.
Exponential growth is characterized by the fact that the rate of change (the growth) becomes increasingly rapid in proportion to the growing total number. In real life, you can see exponential patterns in various processes like population growth and compound interest. Here, by the 30th day, the amount paid reaches a staggering sum compared to static, non-growing payments.
This makes exponential functions incredibly powerful and sometimes deceptive in how quickly they accumulate value. Therefore, exponential functions allow us to understand the potential of growth and how they can rapidly outpace linear (fixed increment) growth.
Exponential growth is characterized by the fact that the rate of change (the growth) becomes increasingly rapid in proportion to the growing total number. In real life, you can see exponential patterns in various processes like population growth and compound interest. Here, by the 30th day, the amount paid reaches a staggering sum compared to static, non-growing payments.
This makes exponential functions incredibly powerful and sometimes deceptive in how quickly they accumulate value. Therefore, exponential functions allow us to understand the potential of growth and how they can rapidly outpace linear (fixed increment) growth.
Payment Calculation
To choose the best payment option in our exercise, we need to thoroughly analyze the calculations involved. First, if we choose option (b), our task involves calculating the total payment received over the span of 30 days. This is done by summing all payments calculated through the geometric series from the previous concept discussion.
Once the sum of payment option (b) is computed using the formula for geometric series, converting the total from cents to dollars simplifies comparison to the fixed \(1,000,000 offered in option (a). For example, the formula yields \(2^{31} - 2\) cents which is equivalent to \)21,474,836.46 after conversion. This mathematically verifies that the exponential payment option (b) significantly surpasses the static one million dollar payout.
Understanding these calculations helps in making informed decisions about salary options, similar patterns of investments, or any real-world scenario where monetary growth is involved.
Once the sum of payment option (b) is computed using the formula for geometric series, converting the total from cents to dollars simplifies comparison to the fixed \(1,000,000 offered in option (a). For example, the formula yields \(2^{31} - 2\) cents which is equivalent to \)21,474,836.46 after conversion. This mathematically verifies that the exponential payment option (b) significantly surpasses the static one million dollar payout.
Understanding these calculations helps in making informed decisions about salary options, similar patterns of investments, or any real-world scenario where monetary growth is involved.
Other exercises in this chapter
Problem 58
Use the Change of Base Formula and a calculator to evaluate the logarithm, rounded to six decimal places. Use either natural or common logarithms. $$\log _{6} 9
View solution Problem 59
Use a graphing device to find all solutions of the equation, rounded to two decimal places. $$\ln x=3-x$$
View solution Problem 59
Use the Change of Base Formula and a calculator to evaluate the logarithm, rounded to six decimal places. Use either natural or common logarithms. $$\log _{7} 2
View solution Problem 60
Use a graphing device to find all solutions of the equation, rounded to two decimal places. $$\log x=x^{2}-2$$
View solution