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Chapter 6: Problem 12

Solve each linear programming problem by the method of corners. $$ \begin{array}{cc} \text { Maximize } & P=3 x-4 y \\ \text { subject to } & x+3 y \leq 15 \\ & 4 x+y \leq 16 \\ & x \geq 0, y \geq 0 \end{array} $$

Short Answer

Expert verified
The maximum value of the linear programming problem is \(P = 45\) at point \((15, 0)\), where \(x = 15\) and \(y = 0\).

Step by step solution

01

Identify the constraints and their graph

First, identify the constraints of the linear programming problem: 1. \(x + 3y \leq 15\) 2. \(4x + y \leq 16\) 3. \(x \geq 0\) 4. \(y \geq 0\) These constraints should be graphed on the coordinate plane.
02

Find the corner points of the feasible region

Next, we will find the corner points of the feasible region, where the given inequalities intersect or touch each other: 1. Intersection of \(x + 3y \leq 15\) and \(4x + y \leq 16\) (Intersection of Constraint 1 and Constraint 2) 2. Intersection of \(x + 3y \leq 15\) and \(x = 0\) (Intersection of Constraint 1 and Constraint 3) 3. Intersection of \(x + 3y \leq 15\) and \(y = 0\) (Intersection of Constraint 1 and Constraint 4) 4. Intersection of \(4x + y \leq 16\) and \(x = 0\) (Intersection of Constraint 2 and Constraint 3) 5. Intersection of \(4x + y \leq 16\) and \(y = 0\) (Intersection of Constraint 2 and Constraint 4)
03

Solve for each corner point

Now, we will solve the system of equations at each intersection point: 1. Intersection of Constraint 1 and Constraint 2: Solve the system of equations: \[ \begin{cases} x + 3y = 15 \\ 4x + y = 16 \end{cases} \] Solving, we get the corner point \((2, 4)\). 2. Intersection of Constraint 1 and Constraint 3: Solve the system of equations: \[ \begin{cases} x + 3y = 15 \\ x = 0 \end{cases} \] Solving, we get the corner point \((0, 5)\). 3. Intersection of Constraint 1 and Constraint 4: Solve the system of equations: \[ \begin{cases} x + 3y = 15 \\ y = 0 \end{cases} \] Solving, we get the corner point \((15, 0)\). 4. Intersection of Constraint 2 and Constraint 3: Solve the system of equations: \[ \begin{cases} 4x + y = 16 \\ x = 0 \end{cases} \] Solving, we get the corner point \((0, 16)\). 5. Intersection of Constraint 2 and Constraint 4: Solve the system of equations: \[ \begin{cases} 4x + y = 16 \\ y = 0 \end{cases} \] Solving, we get the corner point \((4, 0)\).
04

Evaluate the objective function at each corner point

Now, we will evaluate the value of the objective function, \(P = 3x - 4y\), at each corner point: 1. Corner point \((2,4)\): \(P(2,4) = 3(2) - 4(4) = -10\) 2. Corner point \((0,5)\): \(P(0,5) = 3(0) - 4(5) = -20\) 3. Corner point \((15,0)\): \(P(15,0) = 3(15) - 4(0) = 45\) 4. Corner point \((0,16)\): \(P(0,16) = 3(0) - 4(16) = -64\) 5. Corner point \((4,0)\): \(P(4,0) = 3(4) - 4(0) = 12\)
05

Determine the maximum value

From the values of the objective function at each corner point, the maximum value is \(P = 45\) at point \((15, 0)\). Therefore, the solution to the linear programming problem is \(x = 15\) and \(y = 0\), with a maximum value of \(P = 45\).

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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 a critical concept. It represents all the possible combinations of variables that satisfy the given constraints. Imagine it as a shape on a graph, delineated by lines from each of the constraints. These constraints are typically linear equations or inequalities. The feasible region is always on or above the x-axis and y-axis when the constraints involve non-negative values for variables.

In the given exercise, the constraints define a feasible region. This region is formed where all the inequalities intersect on a graph. Each constraint narrows down possible solutions. For example:
  • The inequality \(x + 3y \leq 15\) gives a line on the graph, and we consider the area below this line.
  • The inequality \(4x + y \leq 16\) adds another line, further narrowing down the region.
  • Inequalities \(x \geq 0\) and \(y \geq 0\) ensure that our solutions remain within the first quadrant.
Together, these limits shape the feasible region, which contains all possible solutions to the problem.
Objective Function
The objective function is an integral part of a linear programming problem. It's the expression that needs to be maximized or minimized. In this exercise, the objective function is presented as \(P = 3x - 4y\). This function tells us what we are trying to optimize.

To determine the most optimal solution, we calculate the objective function鈥檚 value at each corner of the feasible region. The resulting values indicate the effectiveness of each possible solution.

For maximizing or minimizing the function, it's crucial to evaluate it at each corner because the maximum or minimum value of a linear function over a convex polygonal region will occur at one of the vertices or corners. The goal is to find which combination of \(x\) and \(y\), within the feasible region, yields the highest or lowest value of \(P\).
Constraints
Constraints in linear programming are rules that define limits on the variables used in the problem. They shape the "feasible region" and determine the bounds within which a solution must lie. Each constraint is an inequality, and together they restrict the domain of the function being optimized.
In our problem, the constraints are:
  • \(x + 3y \leq 15\) 鈥� This means that any solution within the feasible region can鈥檛 exceed this line when combined with the following constraint.
  • \(4x + y \leq 16\) 鈥� Further restricts possible values of \(x\) and \(y\), tightening the feasible region.
  • \(x \geq 0\) and \(y \geq 0\) 鈥� Ensure solutions are not in negative territory since negative amounts do not make sense in most linear programming contexts like production or resource allocation.
These constraints must be adhered to strictly to ensure solutions are not only optimal but also feasible under provided conditions.
Method of Corners
The Method of Corners is a reliable technique applied in linear programming to find the optimal solution of an objective function. Once you have your feasible region, this method allows you to determine which point within that region provides the optimum value (either maximum or minimum) for the objective function.

Here's how it works:
  • Identify all the corner points of the feasible region. These intersections arise where the constraint lines cross each other or the axes.
  • Evaluate the objective function at each corner point. By calculating the function鈥檚 value at these points, you ascertain which point gives the largest or smallest value needed based on your objective (maximize or minimize).
In this problem, the feasibility region was defined by intersection points like (2,4), (0,5), and (15,0). Calculating \(P = 3x - 4y\) at each point then determines the maximum value is 45 at the point (15,0). This method relies on the geometric property that for a linear function, extreme values will occur at the vertices of the polygonal feasible region.

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Most popular questions from this chapter

National Business Machines Corporation manufactures two models of fax machines: \(A\) and \(B\). Each model A costs $$\$ 100$$ to make, and each model B costs $$\$ 150$$. The profits are $$\$ 30$$ for each model-A and $$\$ 40$$ for each model-B fax machine. If the total number of fax machines demanded each month does not exceed 2500 and the company has earmarked no more than $$\$ 600,000 /$$ month for manufacturing costs, find how many units of each model National should make each month in order to maximize its monthly profit. What is the largest monthly profit the company can make?

A farmer plans to plant two crops, A and \(\mathrm{B}\). The cost of cultivating crop \(\mathrm{A}\) is $$\$ 40$$ acre whereas that of crop B is $$\$60/acre$$. The farmer has a maximum of $$\$ 7400$$ available for land cultivation. Each acre of crop A requires 20 labor-hours, and each acre of crop \(\mathrm{B}\) requires 25 labor-hours. The farmer has a maximum of 3300 labor-hours available. If she expects to make a profit of $$\$ 150 $$ acre on crop \(A\) and $$\$ 200$$ acre on crop \(B\), how many acres of each crop should she plant in order to maximize her profit? What is the optimal profit?

Ace Novelty manufactures "Giant Pandas" and "Saint Bernards." Each Panda requires \(1.5 \mathrm{yd}^{2}\) of plush, \(30 \mathrm{ft}^{3}\) of stuffing, and 5 pieces of trim; each Saint Bernard requires \(2 \mathrm{yd}^{2}\) of plush, \(35 \mathrm{ft}^{3}\) of stuffing, and 8 pieces of trim. The profit for each Panda is $$\$ 10$$, and the profit for each Saint Bernard is $$\$ 15$$. If \(3600 \mathrm{yd}^{2}\) of plush, \(66,000 \mathrm{ft}^{3}\) of stuffing, and 13,600 pieces of trim are available, how many of each of the stuffed animals should the company manufacture to maximize its profit? What is the maximum profit?

Construct the dual problem associated with the primal problem. Solve the primal problem. $$ \begin{array}{ll} \text { Minimize } & C=200 x+150 y+120 z \\ \text { subject to } & 20 x+10 y+z \geq 10 \\ & x+y+2 z \geq 20 \\ x & \geq 0, y \geq 0, z \geq 0 \end{array} $$

Solve each linear programming problem by the method of corners. $$ \begin{array}{lr} \text { Maximize } & P=2 x+3 y \\ \text { subject to } & x+y \leq 6 \\ x & \leq 3 \\ x & \geq 0, y & \geq 0 \end{array} $$
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