Question

The cost of producing \(x\) units of a product is given by \(C=x^{3}-15 x^{2}+87 x-73, \quad 4 \leq x \leq 9\) (a) Use a graphing utility to graph the marginal cost function and the average cost function, \(C / x\), in the same viewing window. (b) Find the point of intersection of the graphs of \(d C / d x\) and \(C / x\). Does this point have any significance?

Short answer

The intersection point on graph between the marginal and average cost, obtained by solving \(2x^3 - 15x^2 + 73 = 0\), has significance as it represents the optimal production level where the marginal cost equals the average cost.

Step-by-step solution

Find the Derivative and the Average Cost Function

The derivative of the cost function is found by differentiating \(C = x^3 - 15x^2 + 87x - 73\) to get \(dC/dx = 3x^2 - 30x + 87\). This is the marginal cost function. The average cost function is obtained by dividing the cost function by \(x\), to get \(C/x = x^2 - 15x + 87 - 73/x\).

Graph the Functions

Use a graphing utility to graph the functions. The \(x-\)coordinates on the graph should go from 4 to 9, as per the given range of \(x\).

Find the Intersection Point

To find the intersection point, set \(dC/dx = C/x\) and solve for \(x\). This implies setting \(3x^2 - 30x + 87 = x^2 - 15x + 87 - 73/x\), which can be simplified to \(2x^3 - 15x^2 + 73 = 0\). This equation can either be solved analytically or using the graph to find where the functions intersect.

Interpret the Intersection Point

The intersection point represents the number of product units where the marginal cost equals the average cost. This point has a special economic significance as it represents the optimum production level for the product, where producing additional units would just increase the cost without bringing additional profit.

Key concepts

Marginal Cost

In calculus, marginal cost is a vital concept for understanding how cost functions behave as production levels change.
It is essentially the additional cost incurred when producing one more unit.
Mathematically, the marginal cost is derived as the first derivative of the total cost function.

In the exercise provided, the cost function is given by the equation \(C = x^3 - 15x^2 + 87x - 73\).
To find the marginal cost, we differentiate this function to get \(\frac{dC}{dx} = 3x^2 - 30x + 87\).
This derivative represents how costs increase or decrease as we adjust the production quantity by a single unit.

Understanding and analyzing the marginal cost helps managers and economists:

• Determine the cost-efficiency of producing additional units
• Make informed pricing and production decisions
• Identify economic production scales where costs start to outweigh profits

Average Cost

Average cost is another fundamental concept in cost analysis, providing insights into the cost associated with producing each unit on average. It is determined by dividing the total cost by the quantity produced.
In essence, average cost helps businesses understand the per-unit expense of manufacturing goods within different production levels.

In the given exercise, the average cost function is calculated by \(\frac{C}{x}\), giving us \(x^2 - 15x + 87 - \frac{73}{x}\).
By using this expression, units can be evaluated based on overall production efficiency.

Average cost is beneficial for businesses in several ways:

• It assists in pricing strategies by providing a baseline for minimum price points
• Helps in analyzing the cost-effectiveness of scaling operations
• Offers a comparative measure against marginal costs to determine optimal production levels

Intersection Point

The intersection point between marginal cost and average cost functions on a graph is crucial for determining optimal production levels.
This point is where the cost of producing an additional unit is equal to the average cost of production.

In the exercise, finding the intersection involves setting \(\frac{dC}{dx} = \frac{C}{x}\) and solving the resulting equation \(3x^2 - 30x + 87 = x^2 - 15x + 87 - \frac{73}{x}\).
Simplifying leads us to \(2x^3 - 15x^2 + 73 = 0\).
This equation can be solved either analytically or graphically for the exact \(x\) value.

This point holds special economic significance:

• It indicates the production level at which maximizing efficiency occurs, balancing between costs and profits
• Beyond this point, producing more may not be profitable as costs outweigh unit earnings
• Helps in decision-making, aligning production with demand and cost structures effectively