Problem 81 - Applied Calculus

Question

A company's marginal cost function is \(M C(x)\) (given below), where \(x\) is the number of units. Find the total cost of the first hundred units \((x=0\) to \(x=100)\) $$ M C(x)=6 e^{-0.02 x} $$

Step-by-Step Solution

Answer

The total cost of producing the first 100 units is approximately 259.41.

Step 1: Understand the Problem

We are given the marginal cost function, \(MC(x) = 6e^{-0.02x}\), where \(x\) represents the number of units produced. Our task is to find the total cost of producing the first 100 units, which is determined by integrating the marginal cost from \(x=0\) to \(x=100\).

Step 2: Set Up the Integral

The total cost for producing from \(x=0\) to \(x=100\) can be calculated by integrating the marginal cost function over this interval. Mathematically, this is expressed as: \[C(100) = \int_{0}^{100} 6e^{-0.02x} \, dx\] where \(C(100)\) is the total cost of producing 100 units.

Step 3: Integrate the Function

Integrate the function \(6e^{-0.02x}\) with respect to \(x\). Applying the standard integration rules for exponential functions, we have:\[\int 6e^{-0.02x} \, dx = -300e^{-0.02x} + C\] where \(C\) is the constant of integration.

Step 4: Evaluate the Definite Integral

Now, compute the definite integral from \(x=0\) to \(x=100\):\[C(100) = \left[ -300e^{-0.02x} \right]_{0}^{100} = -300e^{-0.02 \times 100} + 300e^{-0.02 \times 0}\] Simplifying, we find:\[-300e^{-2} + 300(1)\]

Step 5: Calculate the Total Cost

Calculate the numerical values: \(e^{-2} \approx 0.1353\), leading to:\[-300 \times 0.1353 + 300 = -40.59 + 300 = 259.41\] Thus, the total cost of producing the first 100 units is approximately 259.41.

Key Concepts

Integral Calculus

Integral Calculus is a fundamental aspect of calculus that focuses on two main operations: integration and finding antiderivatives. If you're familiar with derivatives, integration can be thought of as the inverse process. While derivatives give us the rate of change, integrals help in finding the total accumulation or the whole quantity from a rate of change.
To understand integral calculus, consider the following key points:

• Definite and Indefinite Integrals: Indefinite integrals represent a family of functions, whereas definite integrals compute a numerical area under the curve over an interval.
• Applications: Common uses include finding areas under curves, volumes, central points, and in various fields like physics, engineering, and economics.
• Notation: The integral sign \(\int\), followed by a function and \(dx\), indicates integration with respect to \(x\).

In the exercise provided, we used integral calculus to accumulate the marginal costs across 100 units, finding the total cost with the help of integration.

Definite Integral

The definite integral is a key concept when it comes to calculating the total value over an interval. It's specific and pinned down by its limits of integration, which dictate where we start and stop when measuring the area under a curve.
Some essential characteristics include:

• Integration Limits: A definite integral is evaluated between two points, \(a\) and \(b\), making notations look like \( \int_{a}^{b} f(x) \, dx \). This yields a specific numerical result rather than a general function.
• Geometric Interpretation: Consider a curve above the \(x\)-axis; the definite integral gives the area between the curve, the \(x\)-axis, and the vertical lines \(x = a\) and \(x = b\).

In the marginal cost exercise, we established integration boundaries from 0 to 100 for the units,evaluating how the marginal costs add up to the total cost.

Exponential Function

An exponential function involves constant increases or decreases in a process, which models a wide range of phenomena in science and economics. These functions are characterized by their base being a constant (often \(e\), approximately equal to 2.718), raised to a variable exponent.
Here's what you should know about exponential functions:

• Form: Expressed as \(f(x) = a \cdot e^{bx}\), where \(a\) and \(b\) are constants, exponential functions can either grow or decay.
• Properties: Exponential growth implies a rapid increase, while decay means a swift decrease, noticeable in populations or radioactive decay.
• Why \(e\)? The base \(e\) gives us the natural exponential function, which is especially useful because its rate of change is proportional to its present value.

In this context, the function \(MC(x) = 6e^{-0.02x}\) represented marginal cost, where \(e^{-0.02x}\) showed an exponential decay reflecting that costs decrease as more units are produced.