Problem 79 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

FUTURE REVENUE A certain oil well that yields 900 barrels of crude oil per month will run dry in 3 years. The price of crude oil is currently \(\$ 92\) per barrel and is expected to rise at the constant rate of 80 cents per barrel per month. If the oil is sold as soon as it is extracted from the ground, how much total future revenue will be obtained from the well?

Step-by-Step Solution

Answer

The total future revenue from the oil well is \$3,434,400.

Step 1: Determine the monthly yield

The oil well yields 900 barrels of crude oil per month.

Step 2: Determine the time frame

The oil well will run dry in 3 years. Since there are 12 months in a year, the total time frame is \[ 3 \text{ years} \times 12 \text{ months/year} = 36 \text{ months} \]

Step 3: Determine the price increase per month

The price of crude oil increases by 80 cents per month. To use this in calculations, convert cents to dollars: \[ 80 \text{ cents} = 0.80 \text{ dollars} \]

Step 4: Find the equation for the monthly oil price

The price per barrel p in month t can be modeled by the equation: \[ p(t) = 92 + 0.80t \] where 92 is the initial price in dollars and 0.80 is the monthly increase in dollars.

Step 5: Calculate the total revenue for each month

The revenue for each month is the product of the number of barrels extracted and the price per barrel. For month t, the revenue R can be determined by: \[ R(t) = 900 \times p(t) = 900 \times (92 + 0.80t) \] which simplifies to: \[ R(t) = 82800 + 720t \]

Step 6: Sum the total revenue over the 36 months

To find the total revenue over the 36 months, sum the revenue function over t from 0 to 35:\[ \text{Total Revenue} = \ \ \sum_{t=0}^{35} (82800 + 720t) \]This breaks down into two sums: \[ \sum_{t=0}^{35} 82800 + \sum_{t=0}^{35} 720t \]The first sum is simply: \[ 82800 \times 36 \]The second sum uses the formula for the sum of the first n integers \( \sum_{t=0}^{n} t = \frac{n(n+1)}{2} \) so: \[ 720 \sum_{t=0}^{35} t = 720 \times \frac{35 \times 36}{2} \]Now calculate each part: \[ 82800 \times 36 = 2,980,800 \]\[ 720 \times 630 = 453,600 \] So the total revenue is: \[ 2,980,800 + 453,600 = 3,434,400 \]

Key Concepts

Applied Calculus

Applied calculus is a powerful tool used to solve real-world problems. In this exercise, we used calculus principles to predict future revenue. Specifically, we modeled price changes and computed total sums over a given time. Calculus helps in economics, engineering, and even biology. One common application is calculating changes over time. Here, we predicted crude oil price increases and used them to find future revenue. Calculus allows for precise predictions and optimizations. Thus, applied calculus ensures businesses like oil companies make informed decisions.

Sum of Arithmetic Series

An arithmetic series is the sum of a sequence of numbers with a constant difference between consecutive terms. In this exercise, the price of oil increases linearly over time. To find the total revenue, we used the sum of an arithmetic series formula. An arithmetic series formula is: \[ S_n = \frac{n(n+1)}{2} \] Where:

• \(n\) is the number of terms.
• \((n+1)\) ensures all terms are included.
• \(S_n\) is the sum of the series.

We calculate monthly revenues and add them together from month 0 to 35. The first part is a sum of constant monthly revenue, multiplied by the total number of months. The second part is the arithmetic series formula, finding the sum of incremental revenue increases. Combining these sums gave us the total future revenue.

Exponential Growth

Exponential growth refers to the increase that becomes more rapid over time. Even though this problem uses linear growth instead of exponential, understanding exponential growth is crucial in applied calculus. Exponential growth occurs often in real-world scenarios like populations or investments. It's defined by a formula: \[ P(t) = P_0 e^{kt} \]Where:

• \(P(t)\) is the amount at time \(t\).
• \(P_0\) is the initial amount.
• \(e\) is the base of the natural logarithm.
• \(k\) is the growth rate.

While the oil price growth in the problem is linear, exponential growth shows much steeper increases, leading to vastly different totals over time. Exponential models are thus more complex but frequently used when evaluating scenarios with compounding growth impacts.