Sal Owns A Clothing Store That Sells Shorts And Graphic T-shirts. He Sells The Shorts For $ 14 \$14 $14 Each And The T-shirts For $ 6 \$6 $6 Each. He Is Limited To The Constraints Shown By The Set Of Inequalities Below. Which Of The Points

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Introduction

Linear programming is a method to achieve the best outcome (such as maximum profit or lowest cost) in a given mathematical model for some list of requirements represented as linear relationships. In this article, we will explore a real-world scenario where Sal owns a clothing store that sells shorts and graphic T-shirts. He sells the shorts for $14\$14 each and the T-shirts for $6\$6 each. He is limited to the constraints shown by the set of inequalities below. We will use linear programming to determine the optimal solution for Sal's clothing store.

The Problem

Sal's clothing store sells two types of products: shorts and graphic T-shirts. The profit from selling each product is as follows:

  • Shorts: $14\$14 each
  • T-shirts: $6\$6 each

However, Sal is limited by the following constraints:

  • The total number of shorts sold cannot exceed 100.
  • The total number of T-shirts sold cannot exceed 200.
  • The total profit from selling shorts and T-shirts cannot exceed $1000\$1000.

We can represent these constraints as a set of inequalities:

  • x≤100x \leq 100 (shorts constraint)
  • y≤200y \leq 200 (T-shirts constraint)
  • 14x+6y≤100014x + 6y \leq 1000 (profit constraint)

where xx is the number of shorts sold and yy is the number of T-shirts sold.

Graphing the Constraints

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

  • The shorts constraint (x≤100x \leq 100) is a vertical line at x=100x = 100.
  • The T-shirts constraint (y≤200y \leq 200) is a horizontal line at y=200y = 200.
  • The profit constraint (14x+6y≤100014x + 6y \leq 1000) is a line with a slope of −14/6=−7/3-14/6 = -7/3 and a yy-intercept of 1000/6=166.671000/6 = 166.67.

Finding the Feasible Region

The feasible region is the area where all the constraints are satisfied. In this case, the feasible region is the area below the profit constraint line, to the left of the shorts constraint line, and below the T-shirts constraint line.

Finding the Optimal Solution

To find the optimal solution, we need to find the point in the feasible region that maximizes the profit. We can do this by finding the intersection of the profit constraint line and the shorts constraint line.

To find the intersection point, we can set the two equations equal to each other:

14x+6y=100014x + 6y = 1000 x=100x = 100

Substituting x=100x = 100 into the first equation, we get:

14(100)+6y=100014(100) + 6y = 1000 1400+6y=10001400 + 6y = 1000 6y=−4006y = -400 y=−66.67y = -66.67

However, this is not a valid solution since yy cannot be negative. Therefore, we need to find another point in the feasible region that maximizes the profit.

Using the Corner Points

To find the optimal solution, we can use the corner points of the feasible region. The corner points are the points where the constraints intersect.

  • The corner point where the shorts constraint and the T-shirts constraint intersect is (100,200)(100, 200).
  • The corner point where the profit constraint and the shorts constraint intersect is (100,−66.67)(100, -66.67).
  • The corner point where the profit constraint and the T-shirts constraint intersect is (−100,200)(-100, 200).

Evaluating the Corner Points

To find the optimal solution, we need to evaluate the profit at each of the corner points.

  • At the corner point (100,200)(100, 200), the profit is: 14(100)+6(200)=1400+1200=260014(100) + 6(200) = 1400 + 1200 = 2600
  • At the corner point (100,−66.67)(100, -66.67), the profit is: 14(100)+6(−66.67)=1400−400=100014(100) + 6(-66.67) = 1400 - 400 = 1000
  • At the corner point (−100,200)(-100, 200), the profit is: 14(−100)+6(200)=−1400+1200=−20014(-100) + 6(200) = -1400 + 1200 = -200

Conclusion

The optimal solution is the corner point (100,200)(100, 200), where the profit is $2600\$2600. This means that Sal should sell 100 shorts and 200 T-shirts to maximize his profit.

Graphical Representation

Here is a graphical representation of the problem:

# Graphical Representation

Feasible Region


The feasible region is the area below the profit constraint line, to the left of the shorts constraint line, and below the T-shirts constraint line.


Corner Points


    

  • Corner point 1: (100,200)(100, 200)
    • 

  • Corner point 2: (100,−66.67)(100, -66.67)
    • 

  • Corner point 3: (−100,200)(-100, 200)
    • 



Optimal Solution


The optimal solution is the corner point (100,200)(100, 200), where the profit is $2600\$2600.


Graph

Mathematical Representation

Here is a mathematical representation of the problem:

# Mathematical Representation

Variables


    

  • xx: number of shorts sold
    • 

  • yy: number of T-shirts sold
    • 



Constraints


    

  • x≤100x \leq 100 (shorts constraint)
    • 

  • y≤200y \leq 200 (T-shirts constraint)
    • 

  • 14x+6y≤100014x + 6y \leq 1000 (profit constraint)
    • 



Objective Function


Maximize: 14x+6y14x + 6y


Solution


The optimal solution is x=100x = 100 and y=200y = 200, which maximizes the profit to $2600\$2600.


Code


import numpy as np

# Define the coefficients of the constraints
A = np.array([[1, 0], [0, 1], [14, 6]])
b = np.array([100, 200, 1000])

# Define the objective function
c = np.array([14, 6])

# Solve the linear programming problem
from scipy.optimize import linprog
res = linprog(c, A_ub=A, b_ub=b)

# Print the optimal solution
print("Optimal solution:", res.x)
print("Optimal profit:", res.fun)
</code></pre>
<h2><strong>Conclusion</strong></h2>
<p>In this article, we used linear programming to solve a real-world problem of Sal's clothing store. We graphed the constraints, found the feasible region, and evaluated the corner points to find the optimal solution. The optimal solution is to sell 100 shorts and 200 T-shirts to maximize the profit to <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">{{content}}lt;/mi><mn>2600</mn></mrow><annotation encoding="application/x-tex">\$2600</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8056em;vertical-align:-0.0556em;"></span><span class="mord">$2600</span></span></span></span>.<br/></p>

<h2><strong>Introduction</strong></h2>
<p>In our previous article, we used linear programming to solve a real-world problem of Sal's clothing store. We graphed the constraints, found the feasible region, and evaluated the corner points to find the optimal solution. In this article, we will answer some frequently asked questions about linear programming and provide additional insights into the problem.</p>
<h2><strong>Q&amp;A</strong></h2>
<h3><strong>Q: What is linear programming?</strong></h3>
<p>A: Linear programming is a method to achieve the best outcome (such as maximum profit or lowest cost) in a given mathematical model for some list of requirements represented as linear relationships.</p>
<h3><strong>Q: What are the constraints in the problem?</strong></h3>
<p>A: The constraints in the problem are:</p>
<ul>
<li>The total number of shorts sold cannot exceed 100.</li>
<li>The total number of T-shirts sold cannot exceed 200.</li>
<li>The total profit from selling shorts and T-shirts cannot exceed <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">{{content}}lt;/mi><mn>1000</mn></mrow><annotation encoding="application/x-tex">\$1000</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8056em;vertical-align:-0.0556em;"></span><span class="mord">$1000</span></span></span></span>.</li>
</ul>
<h3><strong>Q: How do we graph the constraints?</strong></h3>
<p>A: We graph the constraints on a coordinate plane. The <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>x</mi></mrow><annotation encoding="application/x-tex">x</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal">x</span></span></span></span>-axis represents the number of shorts sold, and the <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>y</mi></mrow><annotation encoding="application/x-tex">y</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.625em;vertical-align:-0.1944em;"></span><span class="mord mathnormal" style="margin-right:0.03588em;">y</span></span></span></span>-axis represents the number of T-shirts sold.</p>
<h3><strong>Q: What is the feasible region?</strong></h3>
<p>A: The feasible region is the area where all the constraints are satisfied. In this case, the feasible region is the area below the profit constraint line, to the left of the shorts constraint line, and below the T-shirts constraint line.</p>
<h3><strong>Q: How do we find the optimal solution?</strong></h3>
<p>A: We find the optimal solution by evaluating the corner points of the feasible region. The corner points are the points where the constraints intersect.</p>
<h3><strong>Q: What is the optimal solution?</strong></h3>
<p>A: The optimal solution is to sell 100 shorts and 200 T-shirts to maximize the profit to <span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi mathvariant="normal">{{content}}lt;/mi><mn>2600</mn></mrow><annotation encoding="application/x-tex">\$2600</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8056em;vertical-align:-0.0556em;"></span><span class="mord">$2600</span></span></span></span>.</p>
<h3><strong>Q: How do we use linear programming in real-world problems?</strong></h3>
<p>A: Linear programming can be used in a wide range of real-world problems, such as:</p>
<ul>
<li>Resource allocation</li>
<li>Production planning</li>
<li>Inventory management</li>
<li>Supply chain management</li>
<li>Financial planning</li>
</ul>
<h3><strong>Q: What are the advantages of linear programming?</strong></h3>
<p>A: The advantages of linear programming include:</p>
<ul>
<li>It can be used to solve complex problems with multiple constraints.</li>
<li>It can be used to find the optimal solution in a given mathematical model.</li>
<li>It can be used to evaluate the sensitivity of the solution to changes in the constraints.</li>
</ul>
<h3><strong>Q: What are the limitations of linear programming?</strong></h3>
<p>A: The limitations of linear programming include:</p>
<ul>
<li>It assumes that the relationships between the variables are linear.</li>
<li>It assumes that the constraints are linear.</li>
<li>It may not be able to handle non-linear relationships or non-linear constraints.</li>
</ul>
<h2><strong>Conclusion</strong></h2>
<p>In this article, we answered some frequently asked questions about linear programming and provided additional insights into the problem. We hope that this article has been helpful in understanding the concept of linear programming and its applications in real-world problems.</p>
<h2><strong>Graphical Representation</strong></h2>
<p>Here is a graphical representation of the problem:</p>
<pre><code class="hljs"># Graphical Representation

## Feasible Region

The feasible region is the area below the profit constraint line, to the left of the shorts constraint line, and below the T-shirts constraint line.

## Corner Points

- Corner point 1: $(100, 200)$
- Corner point 2: $(100, -66.67)$
- Corner point 3: $(-100, 200)$

## Optimal Solution

The optimal solution is the corner point $(100, 200)$, where the profit is $\$2600$.

## Graph

</code></pre>
<h2><strong>Mathematical Representation</strong></h2>
<p>Here is a mathematical representation of the problem:</p>
<pre><code class="hljs"># Mathematical Representation

## Variables

- $x$: number of shorts sold
- $y$: number of T-shirts sold

## Constraints

- $x \leq 100$ (shorts constraint)
- $y \leq 200$ (T-shirts constraint)
- $14x + 6y \leq 1000$ (profit constraint)

## Objective Function

Maximize: $14x + 6y$

## Solution

The optimal solution is $x = 100$ and $y = 200$, which maximizes the profit to $\$2600$.

## Code

```python
import numpy as np

# Define the coefficients of the constraints
A = np.array([[1, 0], [0, 1], [14, 6]])
b = np.array([100, 200, 1000])

# Define the objective function
c = np.array([14, 6])

# Solve the linear programming problem
from scipy.optimize import linprog
res = linprog(c, A_ub=A, b_ub=b)

# Print the optimal solution
print(&quot;Optimal solution:&quot;, res.x)
print(&quot;Optimal profit:&quot;, res.fun)
</code></pre>
<h2><strong>Conclusion</strong></h2>
<p>In this article, we used linear programming to solve a real-world problem of Sal's clothing store. We graphed the constraints, found the feasible region, and evaluated the corner points to find the optimal solution. We also answered some frequently asked questions about linear programming and provided additional insights into the problem.</p>