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Chapter 12: Problem 59

The FeatureRich Software Company sells its graphing program, Dogwood, with a volume discount. If a customer buys x copies, then he or she pays $${ }^{17} \$ 500 \sqrt{x}$$. It cost the company 10,000 to develop the program and 2 to manufacture each copy. If a single customer were to buy all the copies of Dogwood, how many copies would the customer have to buy for FeatureRich Software's average profit per copy to be maximized? How are average profit and marginal profit related at this number of copies?

Short Answer

Expert verified
To maximize FeatureRich Software's average profit per copy, a customer should buy 10 copies of Dogwood. At this point, the average profit per copy is higher than the marginal profit, indicating that the average profit has been maximized.

Step by step solution

01

Revenue Function

Given the price function \[P(x) = 17 \cdot 500 \sqrt{x},\] we can find the revenue function by multiplying the price function by the number of copies sold, \(x\). So, the revenue function, denoted as \(R(x)\), is: \[R(x) = xP(x) = 17 \cdot 500x \sqrt{x}.\] 2. Find the cost function, and determine the profit function.
02

Cost and Profit Functions

FeatureRich Software has a fixed cost of 10,000 and a manufacturing cost of 2 per copy. So, the cost function, denoted as \(C(x)\), is: \[C(x) = 10,000 + 2x.\] To find the profit function, subtract the cost function from the revenue function: \[P(x) = R(x) - C(x) = 17 \cdot 500x \sqrt{x} - (10,000 + 2x).\] 3. Find the average profit function.
03

Average Profit Function

The average profit per copy function, denoted as \(A(x)\), is the profit function divided by the number of copies sold, \(x\): \[A(x) = \frac{P(x)}{x} = 17 \cdot 500 \sqrt{x} - \frac{10,000}{x} - 2.\] 4. Find the derivative of the average profit function to determine the extrema.
04

Derivative of the Average Profit Function

The derivative of the average profit function represents the rate of change in the average profit. To find the extrema, which refers to the maximum or minimum value, set the derivative equal to zero, and solve for \(x\): \[A'(x) = \frac{d}{dx} \left(17 \cdot 500 \sqrt{x} - \frac{10,000}{x} - 2\right).\] Upon calculating, we get: \[A'(x) = 4250 - \frac{10,000}{x^2}.\] Setting \(A'(x)\) to zero, we can solve for \(x\): \[4250 - \frac{10,000}{x^2} = 0.\] Solving the equation, we find that \(x = 10\). 5. Find the marginal profit function and compare it to the average profit function.
05

Marginal Profit

The marginal profit is the derivative of the profit function: \[MP(x) = P'(x) = \frac{d}{dx} \left(17 \cdot 500x \sqrt{x} - 10,000 - 2x\right).\] Upon calculating, we get: \[MP(x) = 8,500\sqrt{x} - 2.\] At \(x = 10\), we can find the marginal profit: \[MP(10) = 8,500\sqrt{10} - 2.\] At this point, we can also find the average profit: \[A(10) = 17 \cdot 500\sqrt{10} - \frac{10,000}{10} - 2.\] Comparing the marginal profit and average profit at \(x = 10\), we can say that the average profit is higher as compared to the marginal profit, implying that the average profit per copy is maximized, and the relationship between average and marginal profit is that average profit is higher at this maximum point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Profit Maximization
In business, profit maximization is the process through which a company determines the best output and price levels in order to maximize its return. The core objective is to achieve the highest profit possible given the market structure and other economic factors.

For FeatureRich Software, profit maximization involves finding the sweet spot where the average profit per copy of their Dogwood graphing program is at its peak. The number of copies, x, sold at this optimal point can be determined through a series of calculations that consider the fixed development cost, manufacturing cost per copy, and the volume discount pricing strategy employed by the company.

Using the given revenue and cost functions, the exercise required finding where the average profit function achieved its maximum value. This involved calculating the derivative of the average profit function and setting it to zero to find the critical points which indicate potential maxima. By solving this, we could establish that producing and selling 10 copies would maximize the average profit per copy.
Average Profit Function
The average profit function represents the average amount of profit earned for each unit sold. It is a critical concept for understanding how profits change with varying levels of production or sales.

In FeatureRich Software's situation, the average profit function, noted as A(x), is derived by dividing the profit function P(x) by the number of copies x. This function captures the balance between the total costs (including fixed and variable costs) and the total revenue as more copies are sold.

The average profit calculation provided a clear indicator of profitability per unit, which in this case involved a calculation that took into account the initial development cost spread out over the copies sold as well as the production cost per unit. Understanding and optimizing the average profit function is essential for businesses to determine the best pricing and production strategies to maximize their profits.
Marginal Profit Analysis
Marginal profit analysis focuses on the extra profit generated from selling an additional unit. It is an application of marginal analysis and a cornerstone of economic and business decision-making, because it helps determine the most profitable level of production.

By calculating the marginal profit (MP), companies like FeatureRich Software can evaluate how the profit will change when the number of units produced and sold is incremented by one. The marginal profit is essentially the derivative of the total profit function P(x). For the exercise, when we took the derivative of the profit function to get the MP(x), we could assess how additional sales beyond a certain point would affect profits.

At the maximized average profit point where x = 10, the comparison between marginal profit and average profit reflects the relationship between them. It's a critical juncture; if the marginal profit is less than the average profit at this point, the company knows that producing more units could start to decrease the average profit, suggesting that they've reached the optimal volume for the pricing strategy in place.

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Most popular questions from this chapter

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