Question

Solve each linear programming problem. Maximize \(z=x+3 y\) subject to the constraints \(x \geq 0, \quad y \geq 0, \quad x+y \geq 3, \quad x \leq 5, \quad y \leq 7\)

Short answer

The maximum value of \( z \) is 26 at the point (5, 7).

Step-by-step solution

Identify the Objective Function

The objective function to maximize is given by: \[ z = x + 3y \]

Determine the Constraints

List all the constraints provided in the problem: 1. \( x \geq 0 \) 2. \( y \geq 0 \) 3. \( x + y \geq 3 \) 4. \( x \leq 5 \) 5. \( y \leq 7 \)

Graph the Constraints

Graph each constraint on the coordinate plane: - \( x = 0 \) is the y-axis. - \( y = 0 \) is the x-axis. - \( x + y = 3 \) is a line passing through points (3, 0) and (0, 3). - \( x = 5 \) is a vertical line. - \( y = 7 \) is a horizontal line.

Determine the Feasible Region

The feasible region is the area where all the constraints overlap. This is typically a polygon on the graph.

Identify Corner Points of the Feasible Region

Find the corner points where the lines intersect, as these points will help find the maximum value of the objective function. The relevant points are: - (0, 3) - (3, 0) - (5, 0) - (5, 7) - (0, 7)

Evaluate the Objective Function at Each Corner Point

Calculate the value of \( z = x + 3y \) at each of the identified points: - At (0, 3): \( z = 0 + 3(3) = 9 \) - At (3, 0): \( z = 3 + 3(0) = 3 \) - At (5, 0): \( z = 5 + 3(0) = 5 \) - At (5, 7): \( z = 5 + 3(7) = 26 \) - At (0, 7): \( z = 0 + 3(7) = 21 \)

Select the Maximum Value

Comparing the calculated \( z \) values, the maximum value is 26, which occurs at the point (5, 7).

Key concepts

Objective Function

In linear programming, the objective function is what you aim to maximize or minimize. It is typically expressed as a linear equation. For the given problem, the objective function is: \[ z = x + 3y \]Here, 'z' is the value we want to maximize. The variables 'x' and 'y' represent quantities that can change within the problem's constraints. The coefficients (in this case, 1 for x and 3 for y) indicate how much 'x' and 'y' contribute to the overall value of 'z'.

Constraints

Constraints are conditions that the variables must satisfy. They limit the values that 'x' and 'y' can take. In our problem, the constraints are:

• \( x \geq 0 \)
• \( y \geq 0 \)
• \( x + y \geq 3 \)
• \( x \leq 5 \)
• \( y \leq 7 \)

These constraints ensure that 'x' and 'y' stay within certain boundaries, making the solution feasible. We graph these constraints to visualize the feasible region.

Feasible Region

The feasible region is the area on the graph where all the constraints overlap. This region contains all the possible solutions that satisfy the constraints. When you graph each constraint, you will see lines or boundaries that form a polygon. Inside this polygon is the feasible region. For instance, in our problem, the feasible region is the area where the following intersect:

• The y-axis (\(x = 0\))
• The x-axis (\(y = 0\))
• The line \(x + y = 3\)
• The vertical line \(x = 5\)
• The horizontal line \(y = 7\)

This region is crucial because the maximum or minimum value of the objective function will occur at a point within this area.

Corner Points

In linear programming, the maximum or minimum value of the objective function occurs at one of the corner points (vertices) of the feasible region. To identify these corner points, we look at the intersections of the boundary lines of the constraints. In our problem, the corner points are:

• (0, 3)
• (3, 0)
• (5, 0)
• (5, 7)
• (0, 7)

These points are where the lines defining our constraints meet. We then evaluate the objective function at these points to find the maximum (or minimum) value.

Maximize

To maximize the objective function, we calculate its value at each corner point of the feasible region. Our goal is to find which of these points gives the highest value for 'z'. In our example, the values of 'z' at each corner point are:

• At (0, 3): \( z = 0 + 3(3) = 9 \)
• At (3, 0): \( z = 3 + 3(0) = 3 \)
• At (5, 0): \( z = 5 + 3(0) = 5 \)
• At (5, 7): \( z = 5 + 3(7) = 26 \)
• At (0, 7): \( z = 0 + 3(7) = 21 \)

The maximum value of 'z' is 26, which occurs at the point (5, 7). Therefore, to maximize our objective function \( z = x + 3y \), we focus on the corner point that gives us the highest value within the feasible region.