Problem 8 - The Calculus with Analytic Geometry

Question

If \(C(x)\) dollars is the total cost of producing \(x\) units of a commodity and \(C(x)=2 x^{2}-8 x+18\), find (a) the domain and range of \(C\), (b) the average cost function, (c) the absolute minimum average unit cost, and (d) the marginal cost function. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axes.

Step-by-Step Solution

Answer

(a) Domain: \( x \geq 0 \); Range: \( y \geq 10 \). (b) \( \overline{C}(x) = 2x - 8 + \frac{18}{x} \). (c) Minimum average unit cost: 4 at \( x = 3 \). (d) Marginal cost function: \( C'(x) = 4x - 8 \). (e) Sketch each function on the same axes.

Step 1 - Finding the Domain

The domain of a function includes all the possible values of the independent variable (in this case, x) for which the function is defined. Since we are dealing with a production cost function, we consider only non-negative values of x. Thus, the domain of C(x) is all non-negative real numbers: \ \( \text{Domain: } x \geq 0 \).

Step 2 - Finding the Range

To find the range of the quadratic cost function, determine the vertex of the parabola given by the function \( C(x)=2x^2-8x+18 \). The vertex formula for a quadratic function \( ax^2+bx+c \) is given by \( x = -\frac{b}{2a} \). Here, \( a = 2 \) and \( b = -8 \). So, \( x = -\frac{-8}{2 \times 2} = 2 \).\Next, substitute \( x = 2 \) back into the function to find the y-coordinate: \ \( C(2) = 2(2)^2 - 8(2) + 18 = 2 \cdot 4 - 16 + 18 = 8 - 16 + 18 = 10 \).\ This gives the minimum value of C(x) as 10 when \( x \geq 0 \). Therefore, the range is \( y \geq 10 \).

Step 3 - Finding the Average Cost Function

The average cost function \( \overline{C}(x) \) is found by dividing the total cost function by the number of units produced, x. Therefore, \( \overline{C}(x) = \frac{C(x)}{x} = \frac{2x^2 - 8x + 18}{x} \). Simplify the expression: \( \overline{C}(x) = 2x - 8 + \frac{18}{x} \).

Step 4 - Finding the Absolute Minimum Average Unit Cost

To find the absolute minimum average unit cost, take the derivative of the average cost function \( \overline{C}(x) \) and set it to zero to find critical points: \ \( \overline{C}'(x) = \frac{d}{dx}(2x - 8 + \frac{18}{x}) = 2 - \frac{18}{x^2} \). \ Setting the derivative equal to zero: \ \( 2 - \frac{18}{x^2} = 0 \rightarrow 2 = \frac{18}{x^2} \rightarrow x^2 = 9 \rightarrow x = 3 \).\ Verify that this point is a minimum by checking the second derivative: \ \( \overline{C}''(x) = \frac{d}{dx}(2 - \frac{18}{x^2}) = \frac{36}{x^3} > 0 \text{ for } x > 0 \). \ Thus, \( x = 3 \) is a minimum point. Calculate \( \overline{C}(3) \) to find the minimum average cost: \ \( \overline{C}(3) = 2(3) - 8 + \frac{18}{3} = 6 - 8 + 6 = 4 \). \ Hence, the absolute minimum average unit cost is 4 dollars.

Step 5 - Finding the Marginal Cost Function

The marginal cost function is the derivative of the total cost function \( C(x) \). Therefore, \( C'(x) = \frac{d}{dx}(2x^2 - 8x + 18) = 4x - 8 \).

Step 6 - Drawing the Graphs

Graph the three functions on the same axes: \ 1. Total cost function \( C(x) = 2x^2 - 8x + 18 \). \ 2. Average cost function \( \overline{C}(x) = 2x - 8 + \frac{18}{x} \). \ 3. Marginal cost function \( C'(x) = 4x - 8 \). \ Plot each function over the relevant domain and range, noting that the total cost and marginal cost are quadratic and linear functions, respectively, while the average cost function has a hyperbolic term.

Key Concepts

Domain and Range

Understanding the domain and range of a function is crucial in calculus, especially for cost functions. The domain of a function includes all possible values of the independent variable (in this case, x) for which it is defined. In the context of production costs, only non-negative quantities make sense. Hence, the domain of the given cost function, \(C(x) = 2x^2 - 8x + 18\), is all non-negative real numbers: \(x \geq 0\).

To find the range of the quadratic cost function, we need to identify its vertex. The vertex form of a quadratic function \(ax^2 + bx + c\) is found using the formula \(x = -\frac{b}{2a}\). For \(C(x)\), with \(a = 2\) and \(b = -8\), the vertex occurs at \(x = 2\). Substituting this value back into the function gives the minimum value: \(C(2) = 10\). Therefore, the range of \(C(x)\) is \(y \geq 10\).

Knowing the domain and range helps us understand the bounds within which the cost function operates, making it easier to analyze and predict production costs.

Calculus

Calculus is a powerful tool used to study changes and motion, particularly useful in solving problems related to cost functions. It involves two main branches: differential calculus and integral calculus.

In this exercise, we use differential calculus to determine critical points and derive related functions like the marginal cost and the average cost. By taking derivatives of the total cost function \(C(x)\), we obtain the marginal cost function \(C'(x)\). This derivative represents the rate of change of total cost with respect to the number of units produced.

Understanding the concept of derivatives can reveal insights into the behavior of cost functions, such as finding minimum or maximum points, which are crucial for optimizing production costs.

Marginal Cost Function

The marginal cost function, denoted as \(C'(x)\), represents the additional cost incurred from producing one more unit of the commodity. It's the derivative of the total cost function \(C(x)\).

For the given cost function \(C(x) = 2x^2 - 8x + 18\), we find the marginal cost by taking the derivative: \[ C'(x) = \frac{d}{dx}(2x^2 - 8x + 18) = 4x - 8 \].

This linear function indicates how the cost changes as production increases. Calculating the marginal cost helps businesses make informed decisions about expanding production. When \(C'(x)\) is positive, the costs are increasing with each additional unit, highlighting the importance of maintaining efficiency and managing resources.

Average Cost Function

The average cost function, denoted as \(\overline{C}(x)\), is the total cost divided by the number of units produced. This gives the cost per unit, which is vital for pricing strategies.

For the given cost function, the average cost function is \[ \overline{C}(x) = \frac{C(x)}{x} = \frac{2x^2 - 8x + 18}{x} = 2x - 8 + \frac{18}{x} \].

To find the absolute minimum average unit cost, we take the derivative of the average cost function and set it to zero to find critical points:

• \[ \overline{C}'(x) = \frac{d}{dx}(2x - 8 + \frac{18}{x}) = 2 - \frac{18}{x^2} \]
• Setting \[ \overline{C}'(x) = 0 \]: \[ 2 - \frac{18}{x^2} = 0 \implies x = 3 \]
• Verify with the second derivative: \[ \overline{C}''(x) = \frac{36}{x^3} > 0 \] for \( x > 0 \)

Therefore, the absolute minimum average unit cost occurs at \(x = 3\), giving \(\overline{C}(3) = 4\) dollars.

Understanding the average cost function aids in determining the cost efficiency and pricing strategies for maximizing profit.