Question

Solve each linear programming problem. Minimize \(z=2 x+3 y\) subject to the constraints \(x \geq 0, \quad y \geq 0, \quad x+y \geq 3, \quad x+y \leq 9, \quad x+3 y \geq 6\).

Short answer

The minimum value of \( z \) is 6, occurring at points (3, 0) and (0, 2).

Step-by-step solution

- Identify Feasible Region

Plot the constraints on a graph to identify the feasible region. The inequalities are: 1. \( x \geq 0 \)2. \( y \geq 0 \)3. \( x + y \geq 3 \)4. \( x + y \leq 9 \)5. \( x + 3y \geq 6 \). The feasible region is the area where all these conditions overlap.

- Plot Constraint Inequalities

Draw the lines corresponding to each of the boundary equations: \( x + y = 3, x + y = 9, x + 3y = 6 \). Shade the feasible region where all constraints are satisfied.

- Determine Corner Points

Find the intersection points of the boundary lines to determine the vertices of the feasible region. These points can be found by solving the equations in pairs: 1. Intersection of \( x + y = 3 \) and \( x = 0 \) gives (0, 3).2. Intersection of \( x + y = 3 \) and \( y = 0 \) gives (3, 0).3. Intersection of \( x + y = 9 \) and \( x = 0 \) gives (0, 9).4. Intersection of \( x + y = 9 \) and \( y = 0 \) gives (9, 0).5. Intersection of \( x + 3y = 6 \) and \( x = 0 \) gives (0, 2).6. Intersection of \( x + 3y = 6 \) and \( y = 0 \) gives (6, 0).7. Intersection of \( x + y = 3 \) and \( x + 3y = 6 \) gives (3, 1).8. Intersection of \( x + y = 9 \) and \( x + 3y = 6 \) gives (3, 2).

- Evaluate Objective Function

Substitute these corner points into the objective function \( z = 2x + 3y \): 1. At (0,3):\( z = 2(0) + 3(3) = 9 \)2. At (3,0):\( z = 2(3) + 3(0) = 6 \)3. At (0,9):\( z = 2(0) + 3(9) = 27 \)4. At (9,0):\( z = 2(9) + 3(0) = 18 \)5. At (0,2):\( z = 2(0) + 3(2) = 6 \)6. At (6,0):\( z = 2(6) + 3(0) = 12 \)7. At (3,1):\( z = 2(3) + 3(1) = 9 \)8. At (3,2):\( z = 2(3) + 3(2) = 12 \).

- Find Minimum Value

The minimum value of \( z \) is found by comparing all values calculated in Step 4. The smallest value is \( z = 6 \) at points (3, 0) and (0, 2).

Key concepts

objective function

In linear programming, the objective function is the primary equation you're aiming to maximize or minimize. For this exercise, our objective function is defined as: \( z = 2x + 3y \). This means we're looking to either maximize or minimize the value of \( z \). Every combination of \( x \) and \( y \) in the feasible region will be plugged into this function to see what \( z \) becomes. The goal is to find the smallest (or largest) value possible given the constraints. Let's break it down a bit:

• \( z \) is the value we want to minimize.
• The terms \( 2x \) and \( 3y \) signify how \( x \) and \( y \) contribute to \( z \).
• The objective function provides a rule to determine the goodness of the feasible solutions.

feasible region

The feasible region is the space on a graph where all the constraints of the problem overlap. This is the area where all the inequalities are satisfied simultaneously. To find it, plot each constraint as an equation and shade the region where all constraints intersect. In this exercise, the constraints are:

• \( x \geq 0 \)
• \( y \geq 0 \)
• \( x + y \geq 3 \)
• \( x + y \leq 9 \)
• \( x + 3y \geq 6 \)

The feasible region is the polygon formed where these shaded areas overlap on the graph. It represents all possible combinations of \(x \) and \( y \) that meet all the given conditions. This region will be our focus because only points within it can provide potential solutions to the problem.

constraints

Constraints in linear programming are inequalities that limit the values of the variables. They restrict the feasible region by defining bounds within which the solution must lie. In the given exercise, the constraints are:

• \( x \geq 0 \)
• \( y \geq 0 \)
• \( x + y \geq 3 \)
• \( x + y \leq 9 \)
• \( x + 3y \geq 6 \)

Each constraint lines are drawn on the graph, and the intersections form the boundaries of the feasible region. For example, \( x + y \leq 9 \) means any combination of \( x \) and \( y \) that sums up to 9 or less is acceptable. Constraints shape the feasible region, making some sections of the graph off-limits for potential solutions.

corner points

Corner points, also known as vertices, are critical in linear programming. These are the points where the boundary lines of the feasible region intersect. Here's how you find them:

• Solve pairs of constraint equations together to find their intersection points.
• Check which combinations of \( x \text{ and } y \) satisfy all given constraints.
• In this exercise, the corner points are (3,0), (0,3), (9,0), and others identified by solving pairs of constraints.

After identifying these points, we evaluate the objective function at each one to find the minimum or maximum value. For instance, substituting the corner point (3,0) into our objective equation, \( z = 2(3) + 3(0) = 6 \). Since linear programming problems have their optimal solutions typically at these vertices, examining these points helps pinpoint the minimum or maximum values effectively.