Problem 57 - Thomas Calculus

Question

cost from marginal cost The marginal cost of printing a poster when \(x\) posters have been printed is $$ \frac{d c}{d x}=\frac{1}{2 \sqrt{x}} $$ dollars. Find \(c(100)-c(1),\) the cost of printing posters \(2-100\) .

Step-by-Step Solution

Answer

The cost of printing posters 2 to 100 is 9 dollars.

Step 1: Understand the Problem

We are given the marginal cost function \( \frac{dc}{dx} = \frac{1}{2 \sqrt{x}} \) for printing posters. We need to find the change in cost, \(c(100) - c(1)\), which represents the cost difference between printing from the 2nd to the 100th poster.

Step 2: Identify the Integral

The change in cost \( c(x) \) over the interval [1, 100] can be found by integrating the marginal cost function. This means we should find the integral \( \int \frac{1}{2 \sqrt{x}} \, dx \) to get the cost function \( c(x) + C \).

Step 3: Perform the Integration

To solve \( \int \frac{1}{2 \sqrt{x}} \, dx \), we use the power rule for integration: \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \). Let \( u = x^{1/2} \), so \( du = \frac{1}{2}x^{-1/2} \, dx \). Substituting, we get \( \int du = u + C = x^{1/2} + C \). Thus, \( \int \frac{1}{2 \sqrt{x}} \, dx = \sqrt{x} + C \).

Step 4: Calculate Definite Integral

Assess \( c(100) - c(1) \) by calculating \( \int_{1}^{100} \frac{1}{2 \sqrt{x}} \, dx \). Using the solution from the previous step, the definite integral is \( [\sqrt{x}]_{1}^{100} = \sqrt{100} - \sqrt{1} \).

Step 5: Evaluate the Result

Calculate the expression \( \sqrt{100} - \sqrt{1} \). This is equal to \( 10 - 1 = 9 \). Hence, the cost of printing posters 2 to 100 is 9 dollars.

Key Concepts

Integration

Integration is a fundamental concept in calculus that involves finding the antiderivative or the “integral” of a function. It is used to calculate things like areas under curves, total accumulated quantities, or, as in this case, changes in cost. When we integrate a function, we are summing an infinite number of infinitesimal parts to find a whole sum.

In practice, integration can be represented by the integral sign \(\int\), and it usually involves finding a function called the antiderivative that can describe the accumulation of values. For example, if we know the rate at which costs change (as given by a marginal cost function), integration helps us determine the total change in cost over a specific interval.

• The process requires using different techniques to manipulate the functions into a form that can easily be integrated.
• In our exercise case, we used the power rule for integration, which is one of the fundamental rules applied to simplify and solve integrals effectively.

As we performed integration on the marginal cost function, it translated these instantaneous rates into a concrete change in total cost, making it easier to evaluate how production scale affects total expenditure.

Definite Integral

The definite integral is a specific type of integral that evaluates the accumulation of quantities over a certain interval, yielding a numerical result. Unlike an indefinite integral that results in a general formula plus a constant of integration \(C\), the definite integral is bound by specific upper and lower limits, such as \([a, b]\).

In our exercise, we calculate the definite integral of the marginal cost function \(\int_{1}^{100} \frac{1}{2 \sqrt{x}} \, dx\)\, which summarizes the total change in costs from the 1st to the 100th poster. This involves two main steps:

• Finding the antiderivative (in this case, \(\sqrt{x}\)).
• Applying the Fundamental Theorem of Calculus, which involves evaluating the antiderivative at both endpoints and subtracting the values \([F(b) - F(a)]\).

The definite integral effectively gives us the total cost from producing one poster to producing 100 posters, subtracting off the starting point to find only the cost of producing posters numbered 2 through 100. This process is essential in many fields, from economics to physics, where understanding net changes in systems is crucial.

Marginal Cost Function

The marginal cost function represents how much the cost increases for each additional unit produced. It gives a dynamic view of production costs by describing the cost change rate precisely at any given production level. In economics and business, knowing the marginal cost helps optimize production to ensure resources are used efficiently and profitably.

In our given exercise, the marginal cost function is \(\frac{dc}{dx} = \frac{1}{2 \sqrt{x}}\), where \(x\) represents the number of posters printed. This means that as more posters are printed, the cost for each additional poster changes according to this function. Such functions are valuable for identifying the most cost-effective production levels, avoiding unnecessary expenditure, and maximizing profits.

• The key point is understanding how the marginal cost varies with different levels of production.
• In practice, a firm will often produce up to the point where marginal cost equals marginal revenue for maximized profit.

Analyzing the given marginal cost function through integration provided us with a clear view of how total costs accumulate as more units are produced, connecting the marginal decisions to overall financial planning and strategy.