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Chapter 4: Problem 4

Use the technique developed in this section to solve the minimization problem. $$ \begin{array}{ll} \text { Minimize } & C=x-2 y+z \\ \text { subject to } & x-2 y+3 z \leq 10 \\ & 2 x+y-2 z \leq 15 \\ & 2 x+y+3 z \leq 20 \\ & x \geq 0, y \geq 0, z & \geq 0 \end{array} $$

Short Answer

Expert verified
To minimize the given objective function \(C = x - 2y + z\), subject to the given constraints, we find the vertices of the feasible region: \(A (0, 0, 0)\), \(B (0, 0, \frac{10}{3})\), \(C (0, 5, 5)\), \(D (5, 0, \frac{5}{3})\), and \(E (\frac{5}{3}, \frac{5}{3}, \frac{5}{3})\). After evaluating the objective function at these vertices, we find the minimum value of the objective function is \(-5\), which occurs at vertex \(C\) with \(x = 0\), \(y = 5\), and \(z = 5\).

Step by step solution

01

Identify the objective function and constraints

The objective function we want to minimize is \(C = x - 2y + z\), given the following constraints: 1. \(x - 2y + 3z \leq 10\) 2. \(2x + y - 2z \leq 15\) 3. \(2x + y + 3z \leq 20\) 4. \(x \geq 0\) 5. \(y \geq 0\) 6. \(z \geq 0\)
02

Graph the constraints and find the feasible region

In general, it is difficult to graph a 3-dimensional region. But, to find the vertices of the feasible region, we can find the intersection points of the planes formed by the constraints.
03

Find the vertices of the feasible region

Solve each pair of equations for the intersection points: 1. \(x - 2y + 3z = 10, 2x + y - 2z = 15\) 2. \(x - 2y + 3z = 10, 2x + y + 3z = 20\) 3. \(2x + y - 2z = 15, 2x + y + 3z = 20\) And don't forget the intersections with the coordinate planes: 4. \(x - 2y + 3z = 10, x = 0\) 5. \(2x + y - 2z = 15, x = 0\) 6. \(2x + y + 3z = 20, x = 0\) Using these equations, the vertices of the feasible region are: 1. \(A (0, 0, 0)\) 2. \(B (0, 0, \frac{10}{3})\) 3. \(C (0, 5, 5)\) 4. \(D (5, 0, \frac{5}{3})\) 5. \(E (\frac{5}{3}, \frac{5}{3}, \frac{5}{3})\)
04

Determine the value of the objective function at the vertices of the feasible region

Evaluate the objective function for each of the vertices: 1. \(C_A = 0 - 2(0) + 0 = 0\) 2. \(C_B = 0 - 2(0) + \frac{10}{3} = \frac{10}{3}\) 3. \(C_C = 0 - 2(5) + 5 = -5\) 4. \(C_D = 5 - 2(0) + \frac{5}{3} = \frac{20}{3}\) 5. \(C_E = \frac{5}{3} - 2(\frac{5}{3}) + \frac{5}{3} = 0\)
05

Determine the minimum value of the objective function

The lowest value of the objective function is \(-5\) at the vertex \(C (0, 5, 5)\). Therefore, the minimum value of the objective function is \(-5\), and occurs when \(x = 0\), \(y = 5\), and \(z = 5\).

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Key Concepts

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

Objective Function
In linear programming, the objective function is a mathematical expression that defines what needs to be minimized or maximized in a problem. It's essential for the problem we are looking at since it represents the quantity the decision-makers want to optimize鈥攊n this case, the cost C.

The objective function in the given exercise is defined as C = x - 2y + z. Here, x, y, and z are variables that contribute to the total cost. To solve the problem, we need to find the values of these variables that result in the lowest cost without violating any constraints established by the conditions of the problem.
Feasible Region
The feasible region is the set of all possible points that satisfy the problem's constraints. These constraints are the rules that must be met to ensure the solution is viable. In our exercise, the feasible region is in three dimensions due to the three variables, making direct graphing more challenging.

However, by solving the constraints equations, we can identify certain points known as vertices. These vertices are crucial as they are potential candidates for the minimum or maximum value of the objective function we are trying to optimize. The feasible region is the key to finding the optimal solution because any point outside of it would not fulfill all the restrictions set by the problem's conditions.
Constraints
Constraints are the conditions that provide boundaries for the feasible region. They define the limitations within which the objective function must be optimized. In the problem provided, constraints come from inequalities that involve the variables x, y, and z. These inequalities reflect limitations for the values that these variables can assume.

The constraints include not only the inequalities like x - 2y + 3z 鈮� 10, but also the non-negativity restrictions such as x 鈮� 0. It's essential to handle these correctly because they are often the lines that form the edges of the feasible region. The intersection points of these lines, when plotted, will give the vertices where we evaluate the objective function in search for the optimal solution.
Optimization
Optimization in linear programming is the process of finding the best solution from a set of feasible solutions. After establishing the objective function and feasible region grounded by the constraints, the next step is optimization. For a minimization problem like ours, the goal is to find the lowest value of the objective function C that abides by the given constraints.

To do this, we calculate the value of the objective function at all vertices of the feasible region. The smallest of these values, if it exists, represents the optimum or best, solution. Our exercise identified this minimum value to be -5 at the vertex C (0, 5, 5). This procedure assures us that within the set limits, we have found the most cost-effective solution.

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Most popular questions from this chapter

Use the method of this section to solve each linear programming problem. $$ \begin{aligned} \text { Maximize } & P=x+2 y+3 z \\ \text { subject to } & x+2 y+z \leq 20 \\ & 3 x+y \quad \leq 30 \\ & 2 x+y+z=10 \\ x & \geq 0, y \geq 0, z \geq 0 \end{aligned} $$

Consider the linear programming problem $$ \begin{array}{lr} \text { Maximize } & P=3 x+2 y \\ \text { subject to } & x-y \leq 3 \\ x & \leq 2 \\ & x \geq 0, y \geq 0 \end{array} $$ a. Sketch the feasible set for the linear programming problem. b. Show that the linear programming problem is unbounded. c. Solve the linear programming problem using the simplex method. How does the method break down?

Use the method of this section to solve each linear programming problem. $$ \begin{aligned} \text { Minimize } & C=3 x+2 y+z \\ \text { subject to } & x+2 y+z \leq 20 \\ & 3 x+y \quad \leq 30 \\ & 2 x+y+z=10 \\ x & \geq 0, y \geq 0, z & \geq 0 \end{aligned} $$

Everest Deluxe World Travel has decided to advertise in the Sunday editions of two major newspapers in town. These advertisements are directed at three groups of potential customers. Fach advertisement in newspaper is seen by 70,000 group-A customers, 40,000 group-B customers, and 20,000 group-C customers. Fach advertisement in newspaper II is seen by 10,000 group-A, 20,000 group-B, and 40,000 group-C customers. Fach advertisement in ncwspaper I costs \(\$ 1000\), and cach advertiscment in newspaper II costs \(\$ 800\). Everest would like their advertiscments to be read by at least 2 million pcople from group A. \(1.4\) million people from group \(\mathrm{B}\), and 1 million people from group C. How many advertisements should Everest place in cach newspaper to achicve its advertising goals at a minimum cost? What is the minimum cost?

\(\begin{array}{ll}\text { } & \text { }\end{array}\) Furniture is introducing a new line of executive desks made from a specially selected grade of walnut. Initially, three models \(-\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) -are to be marketed. Each model-A desk requires \(1 \frac{1}{4} \mathrm{hr}\) for fabrication, \(1 \mathrm{hr}\) for assembly, and 1 hr for finishing; each model-B desk requires \(1 \frac{1}{2} \mathrm{hr}\) for fabrication, \(1 \mathrm{hr}\) for assembly, and \(1 \mathrm{hr}\) for finishing; each modelC desk requires \(1 \frac{1}{2} \mathrm{hr}, \frac{3}{4} \mathrm{hr}\), and \(\frac{1}{2} \mathrm{hr}\) for fabrication, assembly, and finishing, respectively. The profit on each model-A desk is \(\$ 26\), the profit on each model-B desk is \(\$ 28\), and the profit on each model-C desk is \(\$ 24\). The total time available in the fabrication department, the assembly department, and the finishing department in the first month of production is \(310 \mathrm{hr}, 205 \mathrm{hr}\), and \(190 \mathrm{hr}\), respectively. To maximize Custom's profit, how many desks of each model should be made in the month? What is the largest profit the company can realize? Are there any resources left over?
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