Problem 46 - Basic Technical Mathematics with Calculus

Question

The total cost \(C\) (in dollars) of production can be interpreted as an area. If the cost per unit \(C^{\prime}\) (in dollars per unit) of producing \(x\) units is given by \(100 /(0.01 x+1)^{2}\), find the total cost of producing 100 units by finding the area under the curve of \(C^{\prime}\) vs \(x\).

Step-by-Step Solution

Answer

The total cost of producing 100 units is approximately $100.

Step 1: Define the Problem

We need to find the total cost of producing 100 units, which is the integral of the cost per unit function, \(C^{\prime}(x) = \frac{100}{(0.01x+1)^2}\), from \(x = 0\) to \(x = 100\).

Step 2: Set Up the Integral

To find the total cost, we integrate the cost per unit function over the interval from 0 to 100. The integral is set up as follows:\[ \int_{0}^{100} \frac{100}{(0.01x+1)^2} \, dx \]

Step 3: Use Substitution

To simplify the integration, use the substitution method. Let \(u = 0.01x + 1\). Then, \(du = 0.01\, dx\) or \(dx = 100 \, du\). Substitute into the integral:\[ \int \frac{100}{u^2} \times 100 \, du = 10000 \int \frac{1}{u^2} \, du \]

Step 4: Integrate

The integral \( \int \frac{1}{u^2} \, du \) is equivalent to \( -\frac{1}{u} + C \). So, the integrated expression becomes:\[ -10000 \left(\frac{1}{u}\right) = -10000 \left(\frac{1}{0.01x + 1}\right) \]

Step 5: Evaluate the Definite Integral

Evaluate the definite integral from 0 to 100:\[ -10000 \left(\frac{1}{1.01}\right) + 10000 \left(\frac{1}{1}\right) \] Calculate the result:\[ = 10000 - \frac{10000}{1.01} \approx 10000 - 9900 = 100 \]

Step 6: Conclusion

The total cost of producing 100 units is found by calculating the area under the curve of the cost per unit function over the specified interval. The total cost comes out to be approximately 100 dollars.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in Integral Calculus, primarily used to calculate the area under a curve within a given interval. This is crucial for problems related to cost functions, where understanding the accumulation of costs over a number of units or time periods can be crucial. In our exercise, the total cost of producing 100 units is represented as the area under the curve of the function \( C'(x) = \frac{100}{(0.01x+1)^2} \).\ When you integrate this function over the interval from \( x = 0 \) to \( x = 100 \), you essentially add up all those cost contributions for each incremental unit produced. The result gives a real-world quantity: the total cost in dollars. The integral from \(0\) to \(100\) captures this accumulation reliably, providing a concrete measure of total cost.

• It represents the accumulation of values, like cost, over a specified range.
• It connects the concept of continuous change back to finite, measurable quantities.

This approach emphasizes how calculus bridges the abstract mathematics of functions and real-world applications, like calculating costs.

Substitution Method

The substitution method is a powerful technique used to simplify integrals, making them easier to evaluate. It's particularly useful when the integral is complex or involves a composite function. By making a strategic substitution, you transform the integral into a simpler, often more recognizable form.In our problem, we used substitution by letting \( u = 0.01x + 1 \). This choice simplified the original function \( \frac{100}{(0.01x+1)^2} \) into a form that's simpler to integrate: \( \frac{1}{u^2} \). Doing so allows us to apply fundamental integration rules more directly.

• Choose a substitution that will simplify the differential and the integral.
• Adjust the differential accordingly (for instance, \( du = 0.01 \, dx \)).

This method helps conceptually as well, turning complex mathematical expressions into forms that often appear in standard integral tables, thus streamlining the process of integration.

Cost Function

The cost function in economics represents the total cost of producing a given quantity of goods. It encompasses both fixed and variable costs, delivering insights into production efficiency. In this particular exercise, understanding how the cost function is expressed mathematically helps us interpret real-world scenarios.Here, the marginal cost function \( C'(x) = \frac{100}{(0.01x+1)^2} \), describes how cost per unit varies with the level of production. It decreases as \( x \) increases, indicating economies of scale: higher production leads to lower per-unit costs.

• Derives from economic principles, showing cost behavior over different production levels.
• Helps identify the total cost by integrating over the production range.

Understanding this function's role in the exercise emphasizes how mathematical models aid in capturing the intricacies of economic activities and production.

Area Under Curve

The concept of the area under the curve is a key application of integral calculus, linking geometrical intuition with analytical mathematics. Calculating the area under a curve like \( C'(x) = \frac{100}{(0.01x+1)^2} \) provides insight into accumulated quantities, such as total cost in this context. Visualizing this concept involves seeing the function graph as a collection of infinitesimally thin rectangles spanning from \( x = 0 \) to \( x = 100 \). When these are summed—integrated—across the interval, they collectively give the total area, equating to the total cost.

• It represents an accumulation, important for variables changing over time or quantity.
• Integral to the solution of real-world problems where total accumulation is needed.

Thus, using the area under the curve connects visual and analytical approaches, reinforcing how calculus applies to modeling and solving practical problems.