Problem 9 - Applied Calculus

Question

Let \(C(q)\) represent the total cost of producing \(q\) items. Suppose \(C(15)=2300\) and \(C^{\prime}(15)=108 .\) Estimate the total cost of producing: (a) 16 items (b) 14 items.

Step-by-Step Solution

Answer

Cost for 16 items: 2408; Cost for 14 items: 2192.

Step 1: Understand the Given Information

We are given that the total cost function is represented by \(C(q)\). We have the following specific values: \(C(15) = 2300\) and the derivative \(C'(15) = 108\). This implies that at \(q = 15\), the cost is \(2300\) and the marginal cost, or the rate of change of cost per unit increase in \(q\), is \(108\).

Step 2: Estimate the Cost for 16 Items Using Linear Approximation

To estimate \(C(16)\), we use the linear approximation formula for small changes based on the derivative: \[ C(q + 1) \approx C(q) + C'(q) \times 1 \]Substituting the given values for \(q = 15\):\[ C(16) \approx C(15) + C'(15) \times 1 = 2300 + 108 \times 1 = 2408 \]

Step 3: Estimate the Cost for 14 Items Using Linear Approximation

To estimate \(C(14)\), similarly use the linear approximation:\[ C(q - 1) \approx C(q) - C'(q) \times 1 \]Substituting the given values for \(q = 15\):\[ C(14) \approx C(15) - C'(15) \times 1 = 2300 - 108 \times 1 = 2192 \]

Step 4: Conclusion

The estimated total cost of producing 16 items is \(2408\) and for 14 items is \(2192\).

Key Concepts

Total Cost Function

In any production process, understanding costs is crucial. The total cost function, often denoted as \( C(q) \), helps in this understanding. It represents the overall cost of producing \( q \) items. This includes all kinds of costs such as raw materials, labor, and overhead.When you break it down:

• Fixed Costs: These costs do not change with the level of output, like rent or salaries.
• Variable Costs: These are costs that vary directly with the quantity of output, like materials and energy.

The total cost function brings together these components, giving a full picture of production expenses at different output levels. In the given exercise, the function \( C(q) \), when evaluated at \( q = 15 \), gives us a total cost of \( 2300 \) units.

Linear Approximation

Linear approximation is a simple yet powerful mathematical technique. It is used to estimate the value of a function near a given point using the function's slope at that point.Imagine you have a point on a curve. If you draw a tangent line at this point, the line will closely approximate the curve for points near the tangent.

• The formula for linear approximation can be simply stated as:
  \[ f(x + \Delta x) \approx f(x) + f'(x) \cdot \Delta x \]
• Here, \( f(x) \) is the function value at the point, \( f'(x) \) is the derivative (or slope) at that point, and \( \Delta x \) is the change in \( x \).

In our problem, this method was applied to estimate the cost of producing 16 and 14 items based on the known values at 15 items. By adjusting \( q \) by ±1 and using the derivative of the cost function, we were able to find close estimates efficiently.

Derivative

Derivatives are central to calculus and are used to measure how a function changes as its input changes. In simpler terms, the derivative of a function gives you the rate at which the function's value changes at any given point.In the context of cost functions:

• The derivative, denoted as \( C'(q) \), represents the marginal cost. It tells us how much the total cost will increase if you produce just one more item.
• For the exercise, \( C'(15) = 108 \) suggests that producing the 16th item increases the cost by 108 units.

Derivatives allow for precise predictions and are essential for optimizing production and cost efficiency. By understanding and using the derivative, businesses can make informed decisions about scaling production up or down.