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Chapter 9: Problem 8

Marketing tells you that if you set the price of an item at $$\$ 10$$ then you will be unable to sell it, but that you can sell 500 items for each dollar below $$\$ 10$$ that you set the price. Suppose your fixed costs total $$\$ 3000$$, and your marginal cost is $$\$ 2$$ per item. What is the most profit you can make?

Short Answer

Expert verified
The optimal price will be determined from the profit function. This is the price that will maximize the profit the business can make on selling a product. The maximum profit corresponds to this optimal price.

Step by step solution

01

Derive the price-quantity relationship

First, determine the relationship between the price and the quantity of units sold. Since 500 items are sold for each dollar under $10, the number of items sold can be represented as Q = 500 * (10 - P), where P is the price per item.
02

Identify Revenue and Total Costs

Revenue (R) for the business will equal the price (P) times the quantity sold (Q) which will translate into R = P * Q. Substitute Q from step 1 into revenue to get a function of revenue in terms of price: R = P*(500*(10-P)) = 5000P - 500P^2. Total costs (TC) will also include fixed costs and the variable costs (marginal cost times quantity), so it can be represented as TC = 3000 + 2Q.
03

Determine Profit

The maximum profit will be revenue subtracted by total costs. Thus, Profit = R - TC. Substitute R and TC from the previous steps to get Profit = (5000P - 500P^2) - (3000 + 2 *(500*(10-P))).
04

Maximizing Profit

Profits are maximized when the derivative of the profit function with respect to price is equal to 0. Differentiate the profit with respect to price and equate to zero to find the price that maximizes profit and then substitute that price back into the profit function obtained in step 3 to compute the actual maximum profit.

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Key Concepts

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

Price-Quantity Relationship
In business, understanding the relationship between the price of a product and the quantity sold is crucial for determining profit. In our exercise, this relationship is defined as selling 500 items for each dollar the price is below \(10. Let's say, for example, if the price is set at \)9, i.e., \(1 less than \)10, you would sell 500 items. Similarly, pricing at \(8, which is \)2 less than $10, results in 1,000 sales. This is given by the formula:
  • \( Q = 500 \times (10 - P) \)
where \( Q \) is the quantity sold, and \( P \) represents the price you set. This tells us how the item price impacts sales volume, which is foundational for revenue and profit calculations.
Revenue Calculation
Revenue is the total income generated from selling goods. To calculate it, multiply the price per item by the number of items sold. Based on the relationship between price and quantity, our revenue formula becomes:
  • \( R = P \times Q \)
  • \( R = P \times (500 \times (10 - P)) \)
  • \( R = 5000P - 500P^2 \)
This formula helps us understand that revenue increases with higher prices and quantities sold, but there is a balance. As price increases, sales volume decreases, affecting revenue. Finding that perfect balance is key to maximizing profits.
Cost Analysis
Costs directly affect profit. In our scenario, fixed costs are \(3000, regardless of sales volume. Then, we have variable costs, computed as the marginal cost of \)2 per item times the quantity sold Calculation of total costs is:
  • \( TC = 3000 + 2Q \)
  • \( TC = 3000 + 2 \times (500 \times (10 - P)) \)
Managing costs efficiently is essential to maximizing profit. Understanding both fixed and variable costs help in deciding how to price items effectively. Reducing costs while optimizing sales can lead to the most profitable operations.
Derivative Calculation
In the quest for profit maximization, derivatives are powerful tools. To find the price that maximizes profit, we examine the derivative of the profit function, set it to zero, and solve for price.First, profit is calculated as revenue minus total costs:
  • \( Profit = R - TC \)
  • \( Profit = (5000P - 500P^2) - (3000 + 2 \times (500 \times (10 - P))) \)
To find where profit peaks, differentiate the profit function with respect to price \( P \), set that derivative equal to zero, and solve for \( P \).This optimal price will offer maximum profit, helping set strategic pricing decisions. By understanding and applying the derivative, businesses can effectively navigate the complex relationship between price, cost, and sales success.

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Most popular questions from this chapter

A box with square base and no top is to hold a volume \(V\). Find (in terms of \(V\) ) the dimensions of the box that requires the least material for the five sides. Also find the ratio of height to side of the base. (This ratio will not involve \(V\).)

A box with square base is to hold a volume \(200 .\) The bottom and top are formed by folding in flaps from all four sides, so that the bottom and top consist of two layers of cardboard. Find the dimensions of the box that requires the least material. Also find the ratio of height to side of the base.

A window consists of a rectangular piece of clear glass with a semicircular piece of colored glass on top. Suppose that the colored glass transmits only \(k\) times as much light per unit area as the clear glass \((k\) is between 0 and 1 ). If the distance from top to bottom (across both the rectangle and the semicircle) is a fixed distance \(H,\) find (in terms of \(k\) ) the ratio of vertical side to horizontal side of the rectangle for which the window lets through the most light.

How are your answers to Problem 8 affected if the cost per item for the \(x\) items, instead of being simply \(\$ 2,\) decreases below \(\$ 2\) in proportion to \(x\) (because of economy of scale and volume discounts) by 1 cent for each 25 items produced?

You are designing a poster to contain a fixed amount \(A\) of printing (measured in square centimeters) and have margins of \(a\) centimeters at the top and bottom and \(b\) centimeters at the sides. Find the ratio of vertical dimension to horizontal dimension of the printed area on the poster if you want to minimize the amount of posterboard needed.
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