Problem 43 - Calculus Early Transcendentals

Question

The cost (in dollars) of producing x units of a certain commodity is \(C(x)=5000+10 x+0.05 x^{2} .\) (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 marginal cost. Its significance will be explained in Section 3.7 )

Step-by-Step Solution

Answer

The average rate of change is 20.25 for \([100, 105]\) and 20.05 for \([100, 101]\); the instantaneous rate at \(x=100\) is 20.

Step 1: Understand the Problem

We are given a cost function \(C(x) = 5000 + 10x + 0.05x^2\). We need to determine the average rate of change of this cost with respect to \(x\) over two different intervals and the instantaneous rate of change at a specific point.

Step 2: Average Rate of Change Formula

The average rate of change of a function \(C(x)\) over an interval \([x_1, x_2]\) is given by \(\frac{C(x_2) - C(x_1)}{x_2 - x_1}\).

Step 3: Calculate Average Rate of Change for Interval [100, 105]

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

Step 4: Calculate Average Rate of Change for Interval [100, 101]

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

Step 5: Find the Instantaneous Rate of Change

The instantaneous rate of change is the derivative \(C'(x)\). Compute this derivative:\[C'(x) = \frac{d}{dx}(5000 + 10x + 0.05x^2) = 10 + 0.1x\]Evaluate \(C'(x)\) at \(x = 100\):\[C'(100) = 10 + 0.1(100) = 10 + 10 = 20\]

Step 6: Interpretation

The average rate of change tells us the average increase in cost per unit increase over an interval, while the instantaneous rate of change at a point gives the change in cost per unit, exactly at that point.

Key Concepts

Marginal Cost

Marginal cost is an important concept in economics because it tells us how the cost of production changes when we produce one more unit of a commodity. This can be particularly useful when planning production levels and pricing strategies.

The marginal cost is effectively the same as finding the instantaneous rate of change of the cost function at a particular production level. This is derived by taking the derivative of the cost function. When you have a cost function like in our exercise, which is expressed as:

• \( C(x) = 5000 + 10x + 0.05x^2 \)

The derivative \( C'(x) = 10 + 0.1x \) gives us the marginal cost. Evaluating this at \( x = 100 \), we find that the marginal cost is 20. This means that producing the 101st unit will add approximately $20 to the total cost. Understanding this allows businesses to make informed decisions about scaling production.

Instantaneous Rate of Change

The instantaneous rate of change is a bit like taking a snapshot of how quickly a function's value is changing at a specific point. It is what gives us a precise look at how, say, costs are varying at a precise level of production.

To calculate this, we look for the derivative of the function in relation to the variable in question—in this case, production level \( x \). This derivative helps us figure out how the function changes for infinitely small changes in \( x \). In practical terms, for our cost function \( C(x) = 5000 + 10x + 0.05x^2 \), the instantaneous rate of change is found by computing:

• \( C'(x) = 10 + 0.1x \)

When \( x = 100 \), this derivative gives us \( C'(100) = 20 \). So, at this level of production, the cost is increasing by $20 per unit—a sharp focus on cost fluctuations at an exact point.

Cost Function

A cost function, such as the one we see in this exercise, allows us to model the total cost associated with production levels. It combines fixed costs with costs that vary depending on how much is being produced.

The function in the problem is given by:

• \( C(x) = 5000 + 10x + 0.05x^2 \)

Here:

• 5000 is the fixed cost, implying it will not change regardless of the amount produced.
• \( 10x \) represents a variable cost that increases linearly with each additional unit produced.
• \( 0.05x^2 \) is a more complex part showing that costs increase at a slightly accelerating rate as production increases—indicating perhaps economies or diseconomies of scale.

By understanding the cost function, businesses can optimize their production levels, identify cost-saving opportunities, and ensure they remain profitable while expanding their operations.