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Chapter 5: Problem 28

Use the given constraints to find the maximum value of the objective function and the ordered pair \((x, y)\) that produces the maximum value. \(x \geq 0, y \geq 0\) \(x+y \leq 20\) \(x+2 y \leq 36\) \(x \leq 14\) a. Maximize: \(z=12 x+15 y\) b. Maximize: \(z=15 x+12 y\)

Short Answer

Expert verified
Maximize for \(12x + 15y\): (14, 1) with z = 198; Maximize for \(15x + 12y\): (14, 1) with z = 222.

Step by step solution

01

- Identify the constraints

List the constraints given in the problem: 1. \(x\geq 0\)2. \(y\geq 0\)3. \(x + y \leq 20\)4. \(x + 2 y \leq 36\)5. \(x \leq 14\)
02

- Graph the constraints

Plot these inequalities on a graph, marking the feasible region where all conditions are satisfied.
03

- Find the corner points of the feasible region

Identify the points of intersection of the constraints' boundary lines within the feasible region. These include the intercepts and intersections of the lines:1. Intersection of \(x=0\) and \(x+2y=36\)2. Intersection of \(y=0\) and \(x+y=20\)3. Intersection of \(x+y=20\) and \(x+2y=36\)4. Intersection of \(x+y=20\) and \(x=14\)5. Intersection of \(x=14\) and \(x+2y=36\)
04

- Evaluate the objective functions at the corner points

Calculate the value of the objective functions at each identified corner point:1. For \((x, y)\): Maximize \(z = 12x + 15y\)2. For \((x, y)\): Maximize \(z = 15x + 12y\)
05

- Determine the maximum values

Find which corner point(s) give the maximum values for each function by comparing calculated values:a. For \(z = 12x + 15y\)b. For \(z = 15x + 12y\)

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Key Concepts

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

Objective Function
The objective function is the mathematical expression you want to maximize or minimize in a linear programming problem. In our exercise, we have two objective functions:
  • First: \(z=12x+15y\)
  • Second: \(z=15x+12y\)

The goal is to find the values of \(x\) and \(y\) that either maximize or minimize these functions. Here, we are looking for solutions that maximize the values. Remember, the objective function relies on the feasible region defined by the constraints.
Feasible Region
The feasible region represents all possible solutions that satisfy the given constraints. In this exercise, the constraints are:
  • \(x \, \textgreater \, 0\)
  • \(y \, \textgreater \, 0\)
  • \(x + y \, \textless \, 20\)
  • \(x + 2y \, \textless \, 36\)
  • \(x \, \textless \, 14\)

Graphically, the feasible region is a polygonal area where all these conditions are true simultaneously. This area is bounded by the intersection points of the lines representing the inequalities.
Graphing Inequalities
Graphing inequalities helps to visualize where the feasible region lies. To start, draw each inequality as if it were an equation. For example, plot \(x + y = 20\) and \(x + 2 y = 36\) as straight lines.
Next, shade the regions that satisfy each inequality. For instance, for \(x + y \, \textless \, 20\), shade the region below the line.
The overlapping shaded area from all inequalities is the feasible region where all constraints hold.
Corner Points Evaluation
Corner points, or vertices, are where the boundary lines intersect and form the limits of the feasible region. To maximize the objective functions, you need to evaluate them at these points. Our key intersections are:
  • Intersection of \(x = 0\) and \(x + 2y = 36\)
  • Intersection of \(y = 0\) and \(x + y = 20\)
  • Intersection of \(x + y = 20\) and \(x + 2y = 36\)
  • Intersection of \(x = 14\) and \(x + y = 20\)
  • Intersection of \(x = 14\) and \(x + 2y = 36\)

Calculate the values of the objective functions at these points to find which one gives the maximum value.
Constraints
Constraints are the limitations or conditions that the solution must satisfy. These are typically presented as inequalities. In our problem, they are:
  • \(x \geq 0\)
  • \(y \geq 0\)
  • \(x + y \leq 20\)
  • \(x + 2 y \leq 36\)
  • \(x \leq 14\)

These inequalities form the boundaries of the feasible region. Any solution outside this region is not valid. Therefore, all evaluation and optimization happen within these constraints.

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