Problem 3 - Finite Mathematics and Applied Calculus

Question

Solve the LP problems. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. Minimize \(\quad c=x+y\) subject to \(\begin{aligned} x+2 y & \geq 6 \\ 2 x+y & \geq 6 \\ x \geq 0, y & \geq 0 . \end{aligned}\)

Step-by-Step Solution

Answer

The optimal solution for the given LP problem is \(x = 0\) and \(y = 0\) with the minimum objective function value \(c = 0\).

Step 1: Understand the objective function and constraints

The given LP problem is to minimize the objective function \(c = x + y\) subject to constraints: 1. \(x + 2y \geq 6\) 2. \(2x + y \geq 6\) 3. \(x \geq 0\) 4. \(y \geq 0\)

Step 2: Find the feasible region

First, graph the linear constraints on a coordinate plane. We will consider the region that satisfies all the constraints: 1. For the constraint \(x + 2y \geq 6\), the line is \(x + 2y = 6\). Draw the line and shade the area above it since it represents \(x + 2y \geq 6\). 2. For the constraint \(2x + y \geq 6\), the line is \(2x + y = 6\). Draw the line and shade the area above it since it represents \(2x + y \geq 6\). 3. For the constraint \(x \geq 0\), shade the area to the right of the y-axis. 4. For the constraint \(y \geq 0\), shade the area above the x-axis. Now, identify the region where all the constraints are satisfied, this is the feasible region.

Step 3: Identify feasible region type

If the feasible region is empty, there will be no optimal solution. If it's unbounded, the objective function can take arbitrary large values. In this case, the feasible region is not empty and is indeed bounded.

Step 4: Find vertices of the feasible region

The vertices of the feasible region can be found by solving the system of equations formed by the intersection of the lines from the constraints. There are four vertices: 1. Intersection of lines \(x + 2y = 6\) and \(2x + y = 6\): Solve this system to get \(x = 2\) and \(y = 2\). 2. Intersection of lines \(x + 2y = 6\) and \(x = 0\): Solve this system to get \(x = 0\) and \(y = 3\). 3. Intersection of lines \(2x + y = 6\) and \(y = 0\): Solve this system to get \(x = 3\) and \(y = 0\). 4. Intersection of line \(x = 0\) and \(y = 0\): The point is the origin \((0, 0)\).

Step 5: Evaluate the objective function at each vertex

To find the optimal solution, calculate the value of the objective function for each vertex: 1. \(c = x + y = 2 + 2 = 4\) at the vertex \((2, 2)\). 2. \(c = x + y = 0 + 3 = 3\) at the vertex \((0, 3)\). 3. \(c = x + y = 3 + 0 = 3\) at the vertex \((3, 0)\). 4. \(c = x + y = 0 + 0 = 0\) at the vertex \((0, 0)\).

Step 6: Find the optimal solution

An optimal solution exists because the feasible region is not empty and is bounded. The minimum value of the objective function occurs at the vertex \((0, 0)\) with a value of \(c=0\). So, the optimal solution is \(x = 0\) and \(y = 0\) with the minimum objective function value \(c = 0\).

Key Concepts

Objective Function

In linear programming, the objective function is a fundamental concept. It represents the quantity that needs to be minimized or maximized. In this exercise, the objective function is defined as \(c = x + y\). This means we are looking to minimize the sum of variables \(x\) and \(y\).

The objective function helps to determine which point in the feasible region will provide the optimal solution. This function can either represent profits that need to be maximized or costs to be minimized. Here, since we aim to minimize \(c\), the objective is to find the smallest possible value of \(x + y\) that still satisfies all the given constraints.

Feasible Region

The feasible region is the area on a graph where all the constraints of a linear programming problem are satisfied simultaneously. To create the feasible region, first plot each constraint as a line on a coordinate plane:

• For \(x + 2y \geq 6\), draw the line \(x + 2y = 6\) and shade above it.
• For \(2x + y \geq 6\), draw the line \(2x + y = 6\) and shade above it.
• For \(x \geq 0\) and \(y \geq 0\), restrict the shading to the first quadrant.

The shaded area fulfilling all constraints is the feasible region. It defines the boundary conditions where the solutions exist, determining which \(x, y\) pairs are valid.

It is important to note that this region can be empty, bounded, or unbounded. In our exercise, the feasible region is not empty and is bounded, meaning there are limitations on both \(x\) and \(y\) from all sides.

Constraints

Constraints are conditions expressed as inequalities that represent the limits within which a solution must be found in a problem. In our linear programming problem, we have the following constraints:

• \(x + 2y \geq 6\)
• \(2x + y \geq 6\)
• \(x \geq 0\)
• \(y \geq 0\)

Constraints form the boundaries of the feasible region. They are crucial because they restrict the set of potential solutions to the linear programming problem. Without constraints, the objective function could be optimized indefinitely. They model real-world limitations, such as available resources, budgets, or time restrictions, making it feasible to solve these problems realistically.

Vertices

Vertices are the points of intersection between the boundaries of the feasible region. Each vertex corresponds to a potential solution for the linear programming problem.
To identify these vertices, calculate where the lines from constraints intersect:

• The intersection of \(x + 2y = 6\) and \(2x + y = 6\) gives the vertex \((2, 2)\).
• The intersection of \(x + 2y = 6\) and \(x = 0\) gives \((0, 3)\).
• The intersection of \(2x + y = 6\) and \(y = 0\) gives \((3, 0)\).
• The intersection of \(x = 0\) and \(y = 0\) is \((0, 0)\).

In linear programming, evaluating the objective function at each vertex helps in determining the optimal solution. Since this problem aims to minimize \(c = x + y\), we calculate the value at each vertex to find the smallest sum, which in this problem occurs at the vertex \((0, 0)\).