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Use graphical methods to solve the following linear programming problem. Minimize: z= 3x + y subject to: x+y≤ 10 2x+2y24 x20, y20 Graph the feasible region using the graphing tool to the right. Click to enlarge graph 14 12- 10- 8- 6- ✔

Use graphical methods to solve the following linear programming problem. Minimize: z= 3x + y subject to: x+y≤ 10 2x+2y24 x20, y20 Graph the feasible region using the graphing tool to the right. Click to enlarge graph 14 12- 10- 8- 6- ✔

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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## Linear Programming Problem: Graphical Method ### Problem Statement Use graphical methods to solve the following linear programming problem. **Minimize:** \( z = 3x + y \) **Subject to:** 1. \( x + y \leq 10 \) 2. \( 2x + 2y \geq 4 \) 3. \( x \geq 0 \), \( y \geq 0 \) --- ### Instructions **Graph the feasible region using the graphing tool to the right.** Click on the "Click to enlarge graph" button to use the graphing tool. ### Explanation of Graph The graph on the right is a Cartesian plane marked with grid lines for detailed plotting. It ranges from 0 to 16 on both the X and Y axes. **Steps to Graph the Feasible Region:** 1. **Plot the inequality \( x + y \leq 10 \):** - Convert the inequality to the equation \( x + y = 10 \). - Plot the line by finding two points. For example, when \( x = 0 \), \( y = 10 \) and when \( y = 0 \), \( x = 10 \). - Shade the region below the line since \( x + y \leq 10 \). 2. **Plot the inequality \( 2x + 2y \geq 4 \):** - Convert the inequality to the equation \( 2x + 2y = 4 \). - Simplify it to \( x + y = 2 \) and plot the line by finding two points. For example, when \( x = 0 \), \( y = 2 \) and when \( y = 0 \), \( x = 2 \). - Shade the region above the line since \( x + y \geq 2 \). 3. **Plot the non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \).** - These constraints indicate the region in the first quadrant (right and above the origin). 4. **Identify the feasible region:** - The feasible region is the area where all the shaded regions overlap. This is where all the constraints are satisfied simultaneously. After plotting, you will see that the feasible region is formed by the intersection of these half-planes. The
Transcribed Image Text:## Linear Programming Problem: Graphical Method ### Problem Statement Use graphical methods to solve the following linear programming problem. **Minimize:** \( z = 3x + y \) **Subject to:** 1. \( x + y \leq 10 \) 2. \( 2x + 2y \geq 4 \) 3. \( x \geq 0 \), \( y \geq 0 \) --- ### Instructions **Graph the feasible region using the graphing tool to the right.** Click on the "Click to enlarge graph" button to use the graphing tool. ### Explanation of Graph The graph on the right is a Cartesian plane marked with grid lines for detailed plotting. It ranges from 0 to 16 on both the X and Y axes. **Steps to Graph the Feasible Region:** 1. **Plot the inequality \( x + y \leq 10 \):** - Convert the inequality to the equation \( x + y = 10 \). - Plot the line by finding two points. For example, when \( x = 0 \), \( y = 10 \) and when \( y = 0 \), \( x = 10 \). - Shade the region below the line since \( x + y \leq 10 \). 2. **Plot the inequality \( 2x + 2y \geq 4 \):** - Convert the inequality to the equation \( 2x + 2y = 4 \). - Simplify it to \( x + y = 2 \) and plot the line by finding two points. For example, when \( x = 0 \), \( y = 2 \) and when \( y = 0 \), \( x = 2 \). - Shade the region above the line since \( x + y \geq 2 \). 3. **Plot the non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \).** - These constraints indicate the region in the first quadrant (right and above the origin). 4. **Identify the feasible region:** - The feasible region is the area where all the shaded regions overlap. This is where all the constraints are satisfied simultaneously. After plotting, you will see that the feasible region is formed by the intersection of these half-planes. The
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