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Chapter 4: Problem 14

Use the method of this section to solve each linear programming problem. $$ \begin{aligned} \text { Maximize } & P=x-2 y+z \\ \text { subject to } & 2 x+3 y+2 z \leq 12 \\ & x+2 y-3 z \geq 6 \\ & x \geq 0, y \geq 0, z \geq 0 \end{aligned} $$

Short Answer

Expert verified
The linear programming problem has an unbounded feasible region. Therefore, the maximum value of the objective function \(P = x - 2y + z\) does not exist.

Step by step solution

01

Identify the feasible region constraints

Make sure you understand the constraints given in the exercise. There are three inequalities forming the constraints of the feasible region: 1. \(2x + 3y + 2z \leq 12\) 2. \(x + 2y - 3z \geq 6\) 3. \(x \geq 0, y \geq 0, z \geq 0\)
02

Rewrite the inequalities into standard form

In order to graph the inequalities, we need to rewrite them in terms of x, y, and z. So, we need to express one variable in terms of the other variables: 1. \(z \leq 6 - x - \frac{3}{2}y\) from \(2x + 3y + 2z \leq 12\) 2. \(z \geq \frac{1}{3}x + \frac{2}{3}y - 2\) from \(x + 2y - 3z \geq 6\) Now we can graph the inequalities.
03

Graph the inequalities in a 3D plane

Graph the inequalities in a 3D plane. You may use software or online graphing calculators to plot the inequalities and find the feasible region where all the constraints hold true. Since the constraints are all linear, the feasible region should be in the form of a polyhedron.
04

Check if the feasible region is bounded or unbounded

Using the graph (either by hand or using a software), check if the feasible region is bounded or unbounded. In this problem, we find that the feasible region is unbounded.
05

Find the vertices of the feasible region

Find the vertices of the feasible region by solving the intersection of the constraint planes. The intersection points are the vertices of the polyhedron. Since the feasible region is boundless, there may not be a unique maximum solution and the maximum value of the objective function might not exist.
06

Evaluate the objective function at each vertex

Calculate the value of the objective function \(P = x - 2y + z\) at each vertex of the feasible region. In this case, since the feasible region is unbounded, we cannot find a unique solution to the problem. The objective function can increase without bound. Conclusion: This problem has an unbounded feasible region and hence the maximum value of the objective function does not exist.

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Key Concepts

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

Feasible Region
In linear programming, the feasible region is the set of all possible points that satisfy the constraints of the problem. These points are typically in the form of a polyhedron or a multidimensional shape, determined by the intersection of planes in space. For this problem, the feasible region is formed by the following constraints:
  • \(2x + 3y + 2z \leq 12\)
  • \(x + 2y - 3z \geq 6\)
  • \(x \geq 0, y \geq 0, z \geq 0\)
To visualize this region, it can be graphed in three dimensions, showing how each constraint limits the possible solutions. The feasible region is essential because it contains all potential solutions to the problem. Knowing whether this region is bounded or unbounded helps determine if a maximum or minimum value of the objective function exists.
Objective Function
The objective function in linear programming is a mathematical expression that you want to optimize鈥攅ither maximize or minimize. In this particular problem, the objective function is given as:
\[P = x - 2y + z\]
This function assigns a numerical value to each coordinate within the feasible region, illustrating the value of each potential solution. Evaluating the function at various points within and at the boundaries of this region is crucial to determine the potential maximum or minimum values. Since the feasible region is unbounded in this case, you may find that the objective function can potentially increase or decrease without limit.
Constraints
Constraints form the essential backbone of a linear programming problem. They define the possible solutions by limiting the values of the variables that satisfy these requirements. In this exercise, the constraints are represented as inequalities, each shaping the feasible region's boundary:
  • \(2x + 3y + 2z \leq 12\)
  • \(x + 2y - 3z \geq 6\)
  • \(x \geq 0, y \geq 0, z \geq 0\)
Each inequality restricts the number of permissible solutions, effectively forming a polyhedron in 3D space. Handling these constraints correctly is critical, as solving these inequalities helps establish the feasible region's vertices and limits.
3D Graphing
Graphing in three dimensions allows you to visualize how multiple constraints interact to form a feasible region. For this linear programming problem, graphing the inequalities shows how each plane intersects and combines to create a polyhedron.
Using either graphing software or a tool like an online calculator, you can plot each constraint, viewing how they impose boundaries on possible solutions. Various tools allow you to rotate and view the graph from different angles, aiding in understanding the feasible region's nature and potential vertices.
Since this problem deals with three variables \(x\), \(y\), and \(z\), 3D graphing becomes a powerful way to comprehend how constraints limit and define the feasible region within the objective function setting.

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Most popular questions from this chapter

Use the method of this section to solve each linear programming problem. $$ \begin{aligned} \text { Minimize } & C=-2 x+3 y \\ \text { subject to } & x+3 y \leq 60 \\ 2 x+y & \geq 45 \\ x & \leq 40 \\ x \geq 0, y & \geq 0 \end{aligned} $$

A company manufactures two products, \(\mathrm{A}\) and \(\mathrm{B}\), on two machines, \(\mathrm{I}\) and II. It has been determined that the company will realize a profit of \(\$$ 3/unit of product A and a profit of \)\$ 4 /\( unit of product \)\mathrm{B}\(. To manufacture 1 unit of product \)\mathrm{A}\( requires 6 min on machine I and 5 min on machine II. To manufacture 1 unit of product \)\mathrm{B}\( requires \)9 \mathrm{~min}\( on machine I and 4 min on machine II. There are \)5 \mathrm{hr}\( of machine time available on machine I and \)3 \mathrm{hr}$ of machine time available on machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit? What is the largest profit the company can realize? Is there any time left unused on the machines?

Ashley has earmarked at most \(\$ 250,000\) for investment in three mutual funds: a money market fund, an international equity fund, and a growth-and- income fund. The money market fund has a rate of return of \(6 \% /\) year, the international equity fund has a rate of return of \(10 \%\) /year, and the growth-andincome fund has a rate of return of \(15 \% /\) year. Ashley has stipulated that no more than \(25 \%\) of her total portfolio should be in the growth-and-income fund and that no more than \(50 \%\) of her total portfolio should be in the international equity fund. To maximize the return on her investment, how much should Ashley invest in each type of fund? What is the maximum return?

\(\begin{array}{l}\text { } & \text { } & \text { }\end{array}\) Pharmaceutical produces three kinds of cold formulas: I. II, and III. It takes \(2.5 \mathrm{hr}\) to produce 1000 bottles of formula I, \(3 \mathrm{hr}\) to produce 1000 bottles of formula II, and \(4 \mathrm{hr}\) to produce 1000 bottles of formula III. The profits for each 1000 bottles of formula I, formula II, and formula III are \(\$ 180, \$ 200\), and \(\$ 300\), respectively. Suppose, for a certain production run, there are enough ingredients on hand to make at most 9000 bottles of formula I, 12,000 bottles of formula II, and 6000 bottles of formula III. Furthermore, suppose the time for the production run is limited to a maximum of \(70 \mathrm{hr}\). How many bottles of each formula should be produced in this production run so that the profit is maximized? What is the maximum profit realizable by the company? Are there any resources left over?

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{aligned} \text { Minimize } & C=12 x+4 y+8 z \\ \text { subject to } & 2 x+4 y+z \geq 6 \\ & 3 x+2 y+2 z=2 \\ & 4 x+y+z \geq 2 \\ & x \geq 0, y \geq 0, z \geq 0 \end{aligned} $$
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