Question

Solve each linear programming problem. Maximize \(z=2 x+y\) subject to the constraints \(x \geq 0, \quad y \geq 0, \quad x+y \leq 6, \quad x+y \geq 1\)

Short answer

The maximum value of z is 11 at the point (5, 1).

Step-by-step solution

- Identify the Objective Function

The objective function we need to maximize is given by: z = 2x + y

- Identify the Constraints

The problem is subject to the following constraints: \( x \geq 0 \), \( y \geq 0 \), \(x+y \leq 6 \), and \( x+y \geq 1 \)

- Graph the Inequalities

Sketch the constraints on a graph: - The line \( x + y = 6 \) divides the plane into two regions. The shaded region will be below this line. - The line \( x + y = 1 \) divides the plane below. The shaded region will be above this line. - The non-negative constraints \( x \geq 0 \) and \( y \geq 0 \) help to keep the solution in the first quadrant.

- Find the Feasible Region

The feasible region is the intersection of all the shaded regions defined by the constraints. This region is bounded by the lines \( x + y = 6 \), \( x + y = 1 \), and the \( x \) and \( y\)-axes.

- Identify Corner Points

Identify the corner points (vertices) of the feasible region. These points are the intersections of the lines: - (0, 1) where \( x + y = 1 \) and \( x = 0 \) - (0, 6) where \( x + y = 6 \) and \( x = 0 \) - (1, 0) where \( x + y = 1 \) and \( y = 0 \) - (5, 1) where \( x + y = 6 \) and \( x + y = 1 \)

- Evaluate the Objective Function at Each Corner Point

Substitute each vertex into the objective function to find the value of \( z = 2x + y \): - At (0, 1): \( z = 2(0) + 1 = 1 \) - At (0, 6): \( z = 2(0) + 6 = 6 \) - At (1, 0): \( z = 2(1) + 0 = 2 \) - At (5, 1): \( z = 2(5) + 1 = 11 \)

- Determine the Maximum Value

The maximum value of \( z \) is the highest value obtained in the evaluations. This occurs at point (5, 1) where \( z = 11 \).

Key concepts

Objective Function

The objective function is a mathematical expression that defines what you want to maximize or minimize in a linear programming problem. In this exercise, the objective function is given by:
\(z = 2x + y\).
Here, the goal is to find the values of \(x\) and \(y\) that will maximize the value of \(z\). Essentially, the objective function serves as your 'target' that you aim to optimize subject to certain limitations called constraints.

Constraints

Constraints are conditions that must be met for the solution to be valid. In our exercise, the constraints are:

• \(x \geq 0\)
• \(y \geq 0\)
• \(x + y \leq 6\)
• \(x + y \geq 1\)

These constraints limit the values that \(x\) and \(y\) can take.
Visualizing these in a graph helps to see where these constraints overlap. This overlap area is crucial as it contains all possible solutions that meet all the constraints.

Feasible Region

The feasible region is the area on the graph where all the constraints overlap. It represents all possible values of \(x\) and \(y\) that satisfy all the given constraints.
To determine the feasible region for our exercise, we plot the constraints on a graph:

• The line \(x + y = 6\) and the area below it.
• The line \(x + y = 1\) and the area above it.
• The non-negative constraints \(x \geq 0\) and \(y \geq 0\), restricting the solution to the first quadrant.

The intersection of these regions gives us the feasible region, which is bounded by the lines \(x + y = 6\), \(x + y = 1\), and the \(x\) and \(y\)-axes.

Corner Points

Corner points, or vertices, are the points where the boundaries of the feasible region intersect. These points are vital to solving a linear programming problem because the maximum or minimum value of the objective function occurs at one of these points.
For this exercise, the corner points are:

• (0, 1) where \(x + y = 1\) and \(x = 0\)
• (0, 6) where \(x + y = 6\) and \(x = 0\)
• (1, 0) where \(x + y = 1\) and \(y = 0\)
• (5, 1) where \(x + y = 6\) and \(x + y = 1\)

You then evaluate the objective function at each of these points to find which one gives the maximum value. In our case, substituting these points into \(z = 2x + y\) shows that the maximum value is 11 at point (5, 1).