Problem 62 - Applied Calculus

Question

A company expects profits of \(60 e^{0.02 t}\) thousand dollars per month, but predicts that if it builds a new and larger factory, its profits will be \(80 e^{0.04 t}\) thousand dollars per month, where \(t\) is the number of months from now. Find the extra profits resulting from the new factory during the first two years \((t=0\) to \(t=24\) ). If the new factory will cost \(\$ 1,000,000\), will this cost be paid off during the first two years?

Step-by-Step Solution

Answer

The extra profit is approximately $2,372,400, which covers the factory cost of $1,000,000.

Step 1: Define the Profit Functions

The current expected profits are given by the function \(60 e^{0.02 t}\) and the profits with a new factory are \(80 e^{0.04 t}\). These represent the monthly profit in thousands of dollars, where \(t\) is the number of months.

Step 2: Determine the Period for Integration

The problem asks for the calculation of extra profits in the first two years, which means from \(t=0\) to \(t=24\) months.

Step 3: Set Up the Profit Difference Integral

The extra profit due to the new factory is the difference between the two profit functions, so we calculate the integral of \(\Delta P(t) = 80 e^{0.04 t} - 60 e^{0.02 t}\) from \(t=0\) to \(t=24\).

Step 4: Calculate the Integral for Extra Profits

The integral to find the total extra profit over the period is:\[\int_{0}^{24} (80 e^{0.04 t} - 60 e^{0.02 t}) \, dt\]The solution involves integrating each term separately:\[\begin{align*}\int (80 e^{0.04 t}) \, dt &= \left[ \frac{80}{0.04} e^{0.04t} \right],\\int (60 e^{0.02 t}) \, dt &= \left[ \frac{60}{0.02} e^{0.02t} \right].\end{align*}\]

Step 5: Evaluate the Integral at t=24 and t=0

First calculate at \(t=24\):\[\begin{align*} P(t=24) &= \left(\frac{80}{0.04} e^{0.04 \times 24} - \frac{60}{0.02} e^{0.02 \times 24}\right), \ &= 2000 e^{0.96} - 3000 e^{0.48}.\end{align*}\]Then at \(t=0\):\[\begin{align*} P(t=0) &= \left(\frac{80}{0.04} e^{0} - \frac{60}{0.02} e^{0}\right), \ &= (2000 - 3000).\end{align*}\]Calculate the difference: extra profit = \[2000 e^{0.96} - 3000 e^{0.48} - (2000 - 3000)\].

Step 6: Numerical Calculation

Using approximations:\[ e^{0.96} \approx 2.6117, \quad e^{0.48} \approx 1.6170 \]The extra profit is:\[ 2000 \times 2.6117 - 3000 \times 1.6170 + 1000 = 5223.4 - 4851 + 1000 = 2372.4 \]Thus, the extra profit over 24 months is approximately $2,372,400.

Step 7: Compare the Extra Profit to Factory Cost

The extra profit $2,372,400 is greater than the cost of the factory, which is $1,000,000. Therefore, the cost of the factory will be recouped in the first two years.

Key Concepts

Profit Functions

In the context of business and economics, profit functions serve as powerful tools that predict the profit generated by a company over time. Specifically, they are mathematical functions that model how profits change with respect to time or other influencing factors. In our given exercise, the profit functions are defined as exponential functions:

• The original profit function is represented by \(60 e^{0.02 t}\), indicating that the expected profits are a function of time \(t\), measured in months. "60" is the initial profit value in thousands of dollars at \(t = 0\).
• The new profit function, if a larger factory is built, is \(80 e^{0.04 t}\). This suggests higher growth in profits over time due to increased capacity or efficiency, starting with an initial value of 80 thousand dollars.

By comparing these functions, businesses can forecast the impact of strategic decisions, such as building a new factory, on future profits.

Integration

Integration is a fundamental concept in calculus used to calculate the total accumulation of quantities over a certain interval. It helps in finding areas under curves, which is especially useful in real-world applications like calculating accumulated profits or costs over time. In this exercise, integration is crucial for determining the extra profit generated by the new factory.When we set up our integration for the extra profit, we define it using the difference between the new and old profit functions:\[\Delta P(t) = \int_{0}^{24} (80 e^{0.04 t} - 60 e^{0.02 t}) \, dt\]This integral calculates the accumulated difference in profits over 24 months (0 to 24), giving the additional revenue captured by building the new factory. The integration of each exponential term requires applying the formula for integrating exponential functions \(\int e^{kt} \, dt = \frac{1}{k} e^{kt}\), where \(k\) is the constant multiplier of the exponent.

Exponential Functions

Exponential functions are a special class of mathematical functions characterized by a constant rate of growth or decay. They are expressed in the form \(a e^{bt}\), where \(a\) is a constant, \(e\) is the base of natural logarithms, and \(b\) is the rate of change parameter. These functions are ubiquitous in modeling situations where change occurs at a rate proportional to the current value, such as population growth, radioactive decay, and, in our case, profit growth.In the present scenario:

• \(60 e^{0.02 t}\) represents a slower growth rate of profit, with \(0.02\) indicating the monthly rate at which the profits expand.
• \(80 e^{0.04 t}\) indicates a faster growth rate, as evidenced by the bigger exponent \(0.04\), suggesting the new factory will produce more significant profit increases over the same period.

Understanding and handling exponential functions is crucial for analyzing scenarios involving rapid changes over time.

Cost Analysis

Cost analysis is the process of evaluating the financial aspects of a decision, ensuring that the benefits justify the expenses. For businesses, cost analysis provides insights on whether investing in new facilities like a factory is fiscally prudent. In our exercise, the factory's cost is given as \\(1,000,000.To discern if this investment pays off:1. Calculate the extra profit during the first two years, resulting from the new factory.2. Compare this profit to the cost of the factory.The integration process reveals an approximate extra profit of \\)2,372,400 over 24 months. Since this profit exceeds the factory's cost, the decision to build the new factory is financially sound. This demonstrates a positive return on investment (ROI), making the construction a profitable business move that not only covers the initial investment but also generates substantial additional revenue.