Problem 14 - Precalculus with Limits

Question

Solving a Linear Programming Problem, sketch the region determined by the constraints. Then find the minimum and maximum values of the objective function (if possible) and where they occur, subject to the indicated constraints. $$ \begin{array}{c}{\text { Objective function: }} \\ {z=5 x+\frac{1}{2} y} \\\ {\text { Constraints: }} \\ {x \geq 0} \\ {y \geq 0} \\ {\frac{1}{2} x+y \leq 8} \\ {x+\frac{1}{2} y \geq 4}\end{array} $$

Step-by-Step Solution

Answer

The minimum value of the objective function is 0 which occurs at (0,0) and the maximum value is 80 which occurs at (16,0).

Step 1: Understanding the constraints and sketching the feasible region

First, we need to sketch out the feasible region defined by the constraints. As we can see, the inequality constraints \(x \geq 0\) and \(y \geq 0\) restrict the region to the first quadrant. The constraints \(\frac{1}{2} x + y \leq 8\) and \(x + \frac{1}{2} y \geq 4\) define a region that forms a parallelogram in the first quadrant. Graph these lines and shade the region that fulfills all of these constraints.

Step 2: Determining the Corner Points

Calculate the corner points of the feasible region by determining where the constraint lines intersect: 1) origin (0,0), 2) intersection of \(x + \frac{1}{2} y = 4\) and \(x = 0\) which gives \((0,8)\), 3) intersection of \(x + \frac{1}{2} y = 4\) and \(\frac{1}{2} x + y = 8\) that gives \((4,4)\), and 4) intersection of \(\frac{1}{2} x + y = 8\) and \(y = 0\) which gives \((16,0)\).

Step 3: Maximizing and Minimizing the Objective Function

Find the maximum and minimum of the objective function by plugging each of the corner points into the objective function \(z=5x+0.5y\). Then, select the highest and lowest values. The values of \(z\) for each corner point are as follows: z(0,0) = 0, z(0,8) = 4, z(4,4) = 22, and z(16,0) = 80. Hence, 0 is the minimum and 80 is the maximum of the objective function.

Key Concepts

Objective Function

In linear programming, the objective function plays a critical role as it represents the main goal of the problem. This goal could involve either maximizing or minimizing certain values.
The objective function is a mathematical expression that can include one or more decision variables. In this particular exercise, our objective function is given by:

• \( z = 5x + \frac{1}{2}y \)

The variables \(x\) and \(y\) need to satisfy certain conditions called constraints. Here, we either wish to find the maximum value or the minimum value of \(z\), depending on the scenario.

• Maximize: If we aim for the greatest possible value, like profits.
• Minimize: If our goal is the smallest possible value, such as costs or losses.

In our exercise, we determined that the maximum value of \(z\) is 80, while the minimum value is 0. This was calculated by evaluating \(z\) at the feasible region's corner points.

Feasible Region

The feasible region in a linear programming problem is a crucial visual representation of all the possible solutions that satisfy the given constraints. Think of it as the search area where possible solutions could exist, often visualized in a graph.
This particular region is determined by the inequalities in a set of constraints. In simple terms, it is the area shaded in the graph that meets all the conditions set by the problem. Here,

• \( x \geq 0 \)
• \( y \geq 0 \)
• \( \frac{1}{2}x + y \leq 8 \)
• \( x + \frac{1}{2}y \geq 4 \)

These constraints restrict the feasible region to a specific polygon shape, called a parallelogram, placed within the first quadrant of a coordinate system. Finding the feasible region is the first step when solving such problems, as it's essential to know where to check for the optimum values of the objective function.

Constraints

Constraints are the mathematical conditions that define the bounds of the feasible region in a linear programming exercise. They are mathematical inequalities or equations involving the decision variables such as \(x\) and \(y\).

• \( x \geq 0 \) and \( y \geq 0 \): These ensure that solutions are non-negative and are in the first quadrant of the graph.
• \( \frac{1}{2}x + y \leq 8 \): This limits the solutions to below a particular mathematical line on the graph.
• \( x + \frac{1}{2}y \geq 4 \): This requires solutions to be above another specific line on the graph.

Together, these constraints define a polygon, more specifically a parallelogram, in which our feasible region lies. Any solution to the linear programming exercise must satisfy all constraints, meaning it must fall within or on the boundaries of this region. If a point does not satisfy even one constraint, it won't be considered within the feasible region, and thus not part of potential solutions.

Corner Points

Corner points are extremely valuable in linear programming as they represent potential optimal solutions. The feasible region is shaped by intersection lines formed by constraints, and each intersection is a corner point. In mathematical problems like ours, it's important to calculate the value of the objective function at these points.First, plot the graph based on constraints, which shows where all these lines intersect. In this exercise, there are four intersections:

• Intersection of \( x + \frac{1}{2}y = 4 \) and \( x = 0 \), gives \((0, 8)\).
• Intersection of \( x + \frac{1}{2}y = 4 \) and \( \frac{1}{2}x + y = 8 \), gives \((4, 4)\).
• Intersection of \( \frac{1}{2}x + y = 8 \) and \( y = 0 \), gives \((16, 0)\).
• Intersection at the origin \((0,0)\).

We calculate the objective function \( z = 5x + \frac{1}{2}y \) at each corner point. Generally, solutions choose the point where \( z \) is maximized or minimized. Here, the minimum value, 0, occurs at (0,0) and the maximum value, 80, occurs at (16,0).