Problem 83 - Calculus and its applications

Question

The total cost, in millions of dollars, for Cheevers, Inc., is given by $$ C(t)=100-50 e^{-t} $$ where \(t\) is the time in years since the start-up date. Find each of the following. a) The marginal cost, \(C^{\prime}(t)\) b) \(C^{\prime}(0)\) c) \(C^{\prime}(4)\) (Round to the nearest thousand.) d) Find \(\lim _{t \rightarrow \infty} C(t)\) and \(\lim _{t \rightarrow \infty} C^{\prime}(t)\).

Step-by-Step Solution

Answer

a) \(C'(t) = 50e^{-t}\); b) \(C'(0) = 50\); c) \(C'(4) \approx 0.001\); d) \(\lim_{t \to \infty} C(t) = 100\), \(\lim_{t \to \infty} C'(t) = 0\).

Step 1: Find the Marginal Cost, C'(t)

To find the marginal cost, we need to differentiate the cost function \(C(t) = 100 - 50e^{-t}\) with respect to \(t\).The derivative of a constant is zero and the derivative of \(-50e^{-t}\) is \(50e^{-t}\) because the derivative of \(e^{-t}\) with respect to \(t\) is \(-e^{-t}\). Multiply by the coefficient -50:\[ C'(t) = 0 + 50e^{-t} = 50e^{-t} \]

Step 2: Calculate C'(0)

Substitute \(t = 0\) into the derivative found in Step 1:\[ C'(0) = 50e^{-0} = 50 \times 1 = 50 \]So, the marginal cost at \(t = 0\) is 50.

Step 3: Calculate C'(4)

Substitute \(t = 4\) into the derivative found in Step 1:\[ C'(4) = 50e^{-4} \]Evaluating the exponential:\( e^{-4} \approx 0.0183 \)\[ C'(4) = 50 \times 0.0183 \approx 0.915 \]Rounding to the nearest thousand, \( C'(4) \approx 0.001 \).

Step 4: Find Limit of C(t) as t approaches infinity

Find \(\lim_{t \to \infty} C(t)\):\[ C(t) = 100 - 50e^{-t} \]As \(t\) approaches infinity, \(e^{-t}\) approaches 0.\[ \lim_{t \to \infty} C(t) = 100 - 50 \times 0 = 100 \]

Step 5: Find Limit of C'(t) as t approaches infinity

Find \(\lim_{t \to \infty} C'(t)\):\[ C'(t) = 50e^{-t} \]As \(t\) approaches infinity, \(e^{-t}\) approaches 0.\[ \lim_{t \to \infty} C'(t) = 50 \times 0 = 0 \]

Key Concepts

Exponential Functions

Exponential functions play a crucial role in mathematics, and they often sneak into real-life applications like finance, population dynamics, and even physics. The core idea here is that exponential functions involve variables as exponents. For example, in the cost function, \[ C(t) = 100 - 50e^{-t} \]we see the term \(e^{-t}\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828. This base is constant but what defines the behavior of the function is the exponent \(-t\).

In our context, the function describes how costs change over time. When \(t\) is small, \(e^{-t}\) yields larger values, resulting in a lower cost because of the subtraction in the formula. As \(t\) grows, the value of \(e^{-t}\) approaches zero very quickly, causing its impact on the total cost to diminish. This is because exponentials decline or grow at rates that far exceed standard rates of increase seen in linear or even polynomial growth, which makes them distinctly powerful in modeling situations where quick growth or decay is observed.

So, understanding the behavior and properties of exponential functions helps us analyze and anticipate changes over time, especially in scenarios where rapid shifts happen, like in technological advancements or resource depletion.

Derivatives

Derivatives are foundational in calculus and are a powerful tool for understanding how a function changes at any given point. Simply put, the derivative of a function at a certain point tells us the rate at which the function's value is changing at that point, essentially capturing the idea of a 'slope' for functions that aren't strictly linear.

For the cost function \[ C(t) = 100 - 50e^{-t} \]we find its derivative \( C'(t) \) to determine the marginal cost. The marginal cost represents the sensitivity of the total cost to incremental changes in time, \(t\).

Finding the derivative involves:

• The derivative of a constant (like 100) is zero since constants do not change.
• For the term involving \(e^{-t}\), we apply the chain rule. The derivative of \(e^{-t}\) is \(-e^{-t}\), and when multiplied by -50 (the term's coefficient), gives us \(50e^{-t}\). Hence, \(C'(t) = 50e^{-t}\).

Knowing the derivative helps us understand how quickly costs are increasing or decreasing at any specific \(t\), which in practical settings, assists businesses in making informed decisions.

Limits

In calculus, limits help us understand the behavior of functions as they approach specific points or infinity. They are key to comprehending trends and potential outcomes in numerous applications.

In our exercise, considering limits lets us examine future projections of costs over time. For instance, the expression \[ \lim_{t \to \infty} C(t) = 100 \] suggests that as time goes on indefinitely, the total cost settles at 100 million dollars. This happens because \(e^{-t}\) decays towards zero as \(t\) increases endlessly, meaning the impact of the exponential decay term vanishes altogether.

Moreover, the limit \[ \lim_{t \to \infty} C'(t) = 0 \] indicates that the rate of change in cost eventually ceases, stabilizing completely. Essentially, at infinite time, costs neither increase nor decrease anymore, which is significant in long-term planning and financial projections.

Understanding limits equips us with the foresight of how systems behave as conditions stretch beyond typical ranges or towards extremes, providing a crucial edge in strategic planning and analysis.