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Optimizing Inventory and Pricing: A Mathematical Approach to John's Clothing Store

John owns a clothing store that sells shorts and graphic T-shirts. As a business owner, he needs to make informed decisions about inventory levels and pricing to maximize profits. However, he is limited by a set of constraints that affect his ability to sell these products. In this article, we will explore the mathematical approach to optimizing inventory and pricing for John's clothing store.

The constraints that John faces are represented by a set of inequalities. These inequalities are:

  • 2s + 3t ≤ 120 (budget constraint)
  • s ≥ 0 (non-negativity constraint for shorts)
  • t ≥ 0 (non-negativity constraint for T-shirts)
  • s ≤ 20 (shorts inventory constraint)
  • t ≤ 30 (T-shirts inventory constraint)

Understanding the Constraints

Let's break down each of the constraints to understand their implications.

  • The budget constraint, 2s + 3t ≤ 120, means that the total revenue from selling shorts and T-shirts cannot exceed $120. This is because each short sells for $12 and each T-shirt sells for $5.
  • The non-negativity constraints, s ≥ 0 and t ≥ 0, mean that John cannot sell a negative number of shorts or T-shirts.
  • The inventory constraints, s ≤ 20 and t ≤ 30, mean that John is limited to selling at most 20 shorts and 30 T-shirts.

The Points of Interest

John is interested in finding the optimal solution that maximizes his profits. To do this, he needs to consider the points of interest, which are the possible combinations of shorts and T-shirts that he can sell. The points of interest are:

  • (0, 0) - no shorts and no T-shirts sold
  • (0, 10) - no shorts sold, 10 T-shirts sold
  • (0, 20) - no shorts sold, 20 T-shirts sold
  • (0, 30) - no shorts sold, 30 T-shirts sold
  • (10, 0) - 10 shorts sold, no T-shirts sold
  • (15, 0) - 15 shorts sold, no T-shirts sold
  • (20, 0) - 20 shorts sold, no T-shirts sold
  • (10, 10) - 10 shorts sold, 10 T-shirts sold
  • (10, 20) - 10 shorts sold, 20 T-shirts sold
  • (10, 30) - 10 shorts sold, 30 T-shirts sold
  • (15, 15) - 15 shorts sold, 15 T-shirts sold
  • (15, 20) - 15 shorts sold, 20 T-shirts sold
  • (15, 30) - 15 shorts sold, 30 T-shirts sold
  • (20, 20) - 20 shorts sold, 20 T-shirts sold
  • (20, 30) - 20 shorts sold, 30 T-shirts sold

Graphing the Constraints

To visualize the constraints, we can graph them on a coordinate plane. The x-axis represents the number of shorts sold, and the y-axis represents the number of T-shirts sold.

  • The budget constraint, 2s + 3t ≤ 120, can be graphed as a line with a slope of -2/3 and a y-intercept of 40.
  • The non-negativity constraints, s ≥ 0 and t ≥ 0, can be graphed as the x-axis and y-axis, respectively.
  • The inventory constraints, s ≤ 20 and t ≤ 30, can be graphed as vertical and horizontal lines, respectively.

Finding the Optimal Solution

To find the optimal solution, we need to find the point of interest that maximizes John's profits. To do this, we can use the following steps:

  1. Graph the constraints on a coordinate plane.
  2. Identify the feasible region, which is the area where all the constraints are satisfied.
  3. Find the point of interest that maximizes John's profits.

The Feasible Region

The feasible region is the area where all the constraints are satisfied. This region is bounded by the budget constraint, the non-negativity constraints, and the inventory constraints.

  • The budget constraint, 2s + 3t ≤ 120, is a line with a slope of -2/3 and a y-intercept of 40.
  • The non-negativity constraints, s ≥ 0 and t ≥ 0, are the x-axis and y-axis, respectively.
  • The inventory constraints, s ≤ 20 and t ≤ 30, are vertical and horizontal lines, respectively.

The Optimal Solution

The optimal solution is the point of interest that maximizes John's profits. This point is located in the feasible region and satisfies all the constraints.

  • The optimal solution is the point (10, 20).
  • This point represents selling 10 shorts and 20 T-shirts.
  • The revenue from selling 10 shorts is $120.
  • The revenue from selling 20 T-shirts is $100.
  • The total revenue is $220.

In this article, we explored the mathematical approach to optimizing inventory and pricing for John's clothing store. We identified the constraints that affect John's ability to sell shorts and T-shirts, graphed the constraints on a coordinate plane, and found the optimal solution that maximizes John's profits. The optimal solution is the point (10, 20), which represents selling 10 shorts and 20 T-shirts. This solution satisfies all the constraints and maximizes John's profits.
Optimizing Inventory and Pricing: A Mathematical Approach to John's Clothing Store - Q&A

In our previous article, we explored the mathematical approach to optimizing inventory and pricing for John's clothing store. We identified the constraints that affect John's ability to sell shorts and T-shirts, graphed the constraints on a coordinate plane, and found the optimal solution that maximizes John's profits. In this article, we will answer some of the most frequently asked questions about optimizing inventory and pricing for John's clothing store.

Q: What are the constraints that affect John's ability to sell shorts and T-shirts?

A: The constraints that affect John's ability to sell shorts and T-shirts are:

  • 2s + 3t ≤ 120 (budget constraint)
  • s ≥ 0 (non-negativity constraint for shorts)
  • t ≥ 0 (non-negativity constraint for T-shirts)
  • s ≤ 20 (shorts inventory constraint)
  • t ≤ 30 (T-shirts inventory constraint)

Q: How do I graph the constraints on a coordinate plane?

A: To graph the constraints on a coordinate plane, you can use the following steps:

  1. Draw the x-axis and y-axis.
  2. Graph the budget constraint, 2s + 3t ≤ 120, as a line with a slope of -2/3 and a y-intercept of 40.
  3. Graph the non-negativity constraints, s ≥ 0 and t ≥ 0, as the x-axis and y-axis, respectively.
  4. Graph the inventory constraints, s ≤ 20 and t ≤ 30, as vertical and horizontal lines, respectively.

Q: How do I find the optimal solution that maximizes John's profits?

A: To find the optimal solution that maximizes John's profits, you can use the following steps:

  1. Graph the constraints on a coordinate plane.
  2. Identify the feasible region, which is the area where all the constraints are satisfied.
  3. Find the point of interest that maximizes John's profits.

Q: What is the optimal solution that maximizes John's profits?

A: The optimal solution that maximizes John's profits is the point (10, 20). This point represents selling 10 shorts and 20 T-shirts.

Q: How do I calculate the revenue from selling shorts and T-shirts?

A: To calculate the revenue from selling shorts and T-shirts, you can use the following steps:

  1. Calculate the revenue from selling shorts: 10 shorts x $12/short = $120.
  2. Calculate the revenue from selling T-shirts: 20 T-shirts x $5/T-shirt = $100.
  3. Calculate the total revenue: $120 + $100 = $220.

Q: What are some common mistakes to avoid when optimizing inventory and pricing?

A: Some common mistakes to avoid when optimizing inventory and pricing include:

  • Not considering all the constraints that affect inventory and pricing.
  • Not graphing the constraints on a coordinate plane.
  • Not finding the optimal solution that maximizes profits.
  • Not calculating the revenue from selling inventory.

Q: How can I apply the mathematical approach to optimizing inventory and pricing to my own business?

A: To apply the mathematical approach to optimizing inventory and pricing to your own business, you can use the following steps:

  1. Identify the constraints that affect inventory and pricing.
  2. Graph the constraints on a coordinate plane.
  3. Find the optimal solution that maximizes profits.
  4. Calculate the revenue from selling inventory.

In this article, we answered some of the most frequently asked questions about optimizing inventory and pricing for John's clothing store. We provided step-by-step instructions on how to graph the constraints on a coordinate plane, find the optimal solution that maximizes profits, and calculate the revenue from selling inventory. By applying the mathematical approach to optimizing inventory and pricing, business owners can make informed decisions that maximize profits and minimize losses.