Problem 51 - Calculus: Early Transcendentals

Question

The cost (in dollars) of producing \( x \) units of a certain commodity is \( C(x) = 5000 + 10x + 0.05x^2 \). (a) Find the average rate of change of \( C \) with respect to \( x \) when the production level is changed (i) from \( x = 100 \) to \( x = 105 \) (ii) from \( x = 100 \) to \( x = 101 \) (b) Find the instantaneous rate of change of \( C \) with respect to \( x \) when \( x = 100 \). ( This is called the \textit{marginal cost}. Its significance will be explained in Section 3.7.)

Step-by-Step Solution

Answer

(a)(i) 20.25, (a)(ii) 20.05, (b) 20

Step 1: Understanding the Average Rate of Change

The average rate of change of the function \( C(x) \) from \( x = a \) to \( x = b \) can be found using the formula \( \frac{C(b) - C(a)}{b - a} \). This formula calculates how much the function \( C \) changes, on average, per unit change in \( x \).

Step 2: Part (a)(i): Calculating from 100 to 105

First, calculate \( C(100) \) and \( C(105) \) using the function \( C(x) = 5000 + 10x + 0.05x^2 \).- \( C(100) = 5000 + 10(100) + 0.05(100)^2 = 5000 + 1000 + 500 = 6500 \)- \( C(105) = 5000 + 10(105) + 0.05(105)^2 = 5000 + 1050 + 551.25 = 6601.25 \)Now, use the average rate of change formula:\[ \frac{C(105) - C(100)}{105 - 100} = \frac{6601.25 - 6500}{5} = \frac{101.25}{5} = 20.25 \]

Step 3: Part (a)(ii): Calculating from 100 to 101

Next, calculate \( C(101) \) using the function.- \( C(101) = 5000 + 10(101) + 0.05(101)^2 = 5000 + 1010 + 510.05 = 6520.05 \)Now, use the average rate of change formula:\[ \frac{C(101) - C(100)}{101 - 100} = \frac{6520.05 - 6500}{1} = 20.05 \]

Step 4: Understanding the Instantaneous Rate of Change

The instantaneous rate of change of \( C(x) \) when \( x = \) 100 is given by the derivative of \( C(x) \), \( C'(x) \). This rate is also known as the marginal cost.

Step 5: Calculating the Derivative of C(x)

To find \( C'(x) \), differentiate the function \( C(x) = 5000 + 10x + 0.05x^2 \).- The derivative of a constant (5000) is 0.- The derivative of \( 10x \) is 10.- The derivative of \( 0.05x^2 \) is \( 0.1x \).Thus, \( C'(x) = 10 + 0.1x \).

Step 6: Part (b): Finding the Marginal Cost at x = 100

Substitute \( x = 100 \) into \( C'(x) \):\[ C'(100) = 10 + 0.1(100) = 10 + 10 = 20 \]The marginal cost when \( x = 100 \) is 20.

Key Concepts

Average rate of change

The average rate of change gives us an insight into how a function behaves between two points. Imagine tracking your speed over a journey. If you cover 100 miles in 2 hours, your average speed is 50 miles per hour, even if your speed varied during the trip.

For a function like cost, represented by \( C(x) \), the average rate of change helps us understand how cost changes on average as production changes. We calculate this by using the formula:

• The change in cost: \( C(b) - C(a) \)
• Divided by the change in production levels: \( b - a \)

This formula essentially tells us the typical increase (or decrease) in cost per unit of production as you move from \( x = a \) to \( x = b \). In our example:

• For \( x = 100 \) to \( x = 105 \), it's \( \frac{6601.25 - 6500}{5} = 20.25 \).
• For \( x = 100 \) to \( x = 101 \), it's \( \frac{6520.05 - 6500}{1} = 20.05 \).

This gives a clear picture of cost fluctuations over specific intervals.

Instantaneous rate of change

The instantaneous rate of change is like the speedometer reading at a particular moment – the exact rate at which something is changing. In calculus, this is given by the derivative of a function at a specific point. For functions representing cost, this rate indicates how cost changes for an infinitesimally small change in production.

Using the derivative of our cost function \( C(x) = 5000 + 10x + 0.05x^2 \), we determine how rapidly the cost is climbing when we precisely add the next unit of production. This is calculated using the function's slope at that exact point which brings us to an important calculus tool called "the derivative".

Marginal cost

In economics, marginal cost refers to the additional cost of producing one more unit of a product. It's highly relevant for producers when deciding on production levels because it informs what happens to the total cost with each additional unit.

Mathematically, marginal cost is the same as the instantaneous rate of change of the cost function, given through its derivative. In our situation, after differentiating the cost function, we substitute \( x = 100 \) into the derivative, \( C'(x) = 10 + 0.1x \), to get the marginal cost:

• \( C'(100) = 10 + 0.1(100) = 20 \)

This tells manufacturers that at production level 100, the cost to produce one more unit will increase by 20 dollars.

Derivative

The derivative is a cornerstone concept in calculus, representing how a function changes at any point. It is the mathematical tool that allows us to compute the instantaneous rate of change.

To find the derivative of a function, we apply differentiation rules:

• The derivative of constant terms, like 5000, is 0.
• For linear terms, such as \( 10x \), the coefficient is the derivative.
• For quadratic terms, like \( 0.05x^2 \), you multiply by the exponent (2 in this case), then reduce the exponent by one.

Therefore, differentiating \( C(x) = 5000 + 10x + 0.05x^2 \) results in \( C'(x) = 10 + 0.1x \). This comprehensive expression tells us how the full function behaves as \( x \) changes.

Quadratic function

A quadratic function is characterized by its highest power being 2, typically written in the format \( ax^2 + bx + c \). This type of function creates a parabolic curve when graphed, which can open upwards or downwards depending on the parameters.

In our cost function \( C(x) = 5000 + 10x + 0.05x^2 \), it's quadratic due to the \( 0.05x^2 \) term. Such functions are significant in diverse applications because they elegantly model phenomena with acceleration, like the costs that increase at an increasing rate due to economies of scale or production constraints.

• The vertex of the parabola represents the function's maximum or minimum cost.
• The axis of symmetry provides the line that splits the parabola into two mirror-image halves.

Understanding quadratic functions enhances a student's ability to predict how outputs shift in response to various inputs in both physical and economical contexts.