Problem 84 - Calculus and its applications

Question

The total cost, in millions of dollars, for Marcotte Industries is given by $$ C(t)=200-40 e^{-t} $$ where \(t\) is the time in years since the start-up date. Find each of the following. a) The marginal \(\operatorname{cost} C^{\prime}(t)\) b) \(C^{\prime}(1)\) c) \(C^{\prime}(5)\) (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) \(C'(t) = 40e^{-t}\); b) \(C'(1) \approx 14.716\); c) \(C'(5) \approx 0.270\); d) \(\lim_{t \to \infty} C(t) = 200\), \(\lim_{t \to \infty} C'(t) = 0\).

Step 1: Differentiate the Cost Function

The marginal cost function is the derivative of the cost function, \(C(t)\). We need to find \(C'(t)\). The given function is \(C(t)=200-40e^{-t}\). To differentiate, apply the derivative rules: the derivative of a constant is zero, and for the term \(-40e^{-t}\), use the chain rule. Hence, \(C'(t) = 0 + 40e^{-t}\) because the derivative of \(e^{-t}\) is \(-e^{-t}\). Thus, \(C'(t) = 40e^{-t}\).

Step 2: Calculate the Marginal Cost at t=1

Now, we calculate \(C'(1)\) by substituting \(t = 1\) into the derivative. Therefore, \(C'(1) = 40e^{-1}\). Using a calculator, approximate \(e^{-1} \approx 0.3679\). Thus, \(C'(1) \approx 40 0.3679 = 14.716\) million dollars.

Step 3: Calculate the Marginal Cost at t=5

Next, calculate \(C'(5)\) by substituting \(t = 5\) into the derivative. Thus, \(C'(5) = 40e^{-5}\). Approximate \(e^{-5} \approx 0.00674\). Therefore, \(C'(5) \approx 40 0.00674 \approx 0.2696\) million dollars. Round this to the nearest thousand, which is \(0.270\) million dollars.

Step 4: Evaluate the Limit of C(t) as t Approaches Infinity

As \(t\) approaches infinity, the exponential term \(e^{-t}\) approaches zero. Therefore, \(C(t) = 200 - 40e^{-t}\) approaches \(200\). Thus, \(\lim_{t \to \infty} C(t) = 200\) million dollars.

Step 5: Evaluate the Limit of C'(t) as t Approaches Infinity

For \(C'(t) = 40e^{-t}\), as \(t\) approaches infinity, \(e^{-t}\) approaches zero. Therefore, \(C'(t)\) approaches \(0\). Hence, \(\lim_{t \to \infty} C'(t) = 0\) million dollars.

Key Concepts

Marginal Cost

The concept of marginal cost is crucial in understanding how the cost of production changes with an additional unit of output. In simple terms, it's the cost of producing one more item. Marginal cost is derived by taking the derivative of the total cost function with respect to time or quantity. In this exercise, the total cost function is given as:\[ C(t) = 200 - 40 e^{-t} \]To find the marginal cost, you differentiate the cost function, giving:\[ C'(t) = \frac{d}{dt}(200 - 40e^{-t}) = 40e^{-t} \]This represents the instantaneous rate of change of cost as production changes over time. Understanding marginal costs helps businesses decide if increasing production is financially viable.

Exponential Functions

Exponential functions are mathematical functions of the form \( f(x) = a \cdot e^{bx} \), where \( e \) is the base of natural logarithms, approximately equal to 2.71828. These functions model a multitude of real-world scenarios, such as population growth, radioactive decay, and in this case, the cost changes over time.In our example, the component \[ -40e^{-t} \]is an exponential function. As time \( t \) increases, the term \( e^{-t} \) decreases rapidly towards zero. This diminishing effect explains why costs asymptotically approach a certain value over time as seen when evaluating limits. Exponential functions are key in calculus for modeling scenarios that change at a rate proportional to their size.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a quantity changes. Imagine you're driving a car—the derivative would be similar to the speedometer, showing how quickly you're changing your position.To find the derivative of a function, like our cost function, you're essentially finding the slope of the tangent line to the curve at any point \( t \). For the given cost function:\[ C'(t) = 40e^{-t} \]This tells us how the cost changes over time. The derivative utilizes rules such as the chain rule and derivatives of exponential functions, both essential tools in calculus. Mastering derivatives opens up a world of understanding about change rates in various contexts.

Limits

Limits in calculus refer to the value that a function approaches as the input approaches some value. They form the basis for defining derivatives and integrals and help tackle problems involving infinite behavior.When we evaluated the limit of the cost function:\[ \lim_{t \to \infty} C(t) = 200 \]we were finding the behavior of the cost as time goes on indefinitely. Additionally, for the derivative function:\[ \lim_{t \to \infty} C'(t) = 0 \]This suggests that as time goes forward, the marginal cost becomes negligible. Understanding limits is crucial for predicting the long-term behavior of functions and forms the stepping stone to more advanced calculus concepts like continuity and the definition of derivatives.