Question

Solve each linear programming problem. Minimize \(z=5 x+4 y\) subject to the constraints \(x \geq 0, \quad y \geq 0, \quad x+y \geq 2, \quad 2 x+3 y \leq 12, \quad 3 x+y \leq 12\)

Short answer

The minimum value is 8 at the point \((0, 2)\).

Step-by-step solution

Identify the Objective Function

The objective is to minimize the function given by: \( z = 5x + 4y \)

List the Constraints

The constraints for the problem are: \( x \geq 0 \) \( y \geq 0 \) \( x + y \geq 2 \) \( 2x + 3y \leq 12 \) \( 3x + y \leq 12 \)

Graph the Feasible Region

Graph the lines for each inequality on a coordinate plane: 1. \( x = 0 \) (vertical line) 2. \( y = 0 \) (horizontal line) 3. \( x + y = 2 \) 4. \( 2x + 3y = 12 \) 5. \( 3x + y = 12 \) Then, identify the region that satisfies all these inequalities. This region is bounded by the lines and includes all intersection points.

Determine the Corner Points of the Feasible Region

Find the intersection points of the boundary lines: 1. \( (0, 2) \) by solving \( x + y = 2 \) and \( x = 0 \) 2. \( (2, 0) \) by solving \( x + y = 2 \) and \( y = 0 \) 3. \( (3, 0) \) by solving \( 3x + y = 12 \) and \( y = 0 \) 4. \( (0, 4) \) by solving \( 2x + 3y = 12 \) and \( x = 0 \) 5. \( (3, 3) \) by solving \( 2x + 3y = 12 \) and \( 3x + y = 12 \)

Evaluate the Objective Function at Each Corner Point

Substitute each corner point into the objective function \( z = 5x + 4y \): 1. At \((0, 2)\): \( z = 5(0) + 4(2) = 8 \) 2. At \((2, 0)\): \( z = 5(2) + 4(0) = 10 \) 3. At \((3, 0)\): \( z = 5(3) + 4(0) = 15 \) 4. At \((0, 4)\): \( z = 5(0) + 4(4) = 16 \) 5. At \((3, 3)\): \( z = 5(3) + 4(3) = 27 \)

Select the Minimum Value

Compare the values from the previous step. The smallest value of \( z \) is 8, which occurs at the point \((0, 2)\).

Key concepts

Objective Function

In linear programming, the objective function is the mathematical expression we aim to maximize or minimize.
In this problem, the objective function is given as \( z = 5x + 4y \).
Our goal is to find the values of \(x\) and \(y\) that will minimize this function. This function represents the quantity we seek to optimize, whether it's cost, profit, or another measurable goal.
By changing the values of \(x\) and \(y\) within the feasible region defined by the constraints, we look for that 'optimal' point where \(z\) is minimal.

Feasible Region

The feasible region is the area on the graph where all the constraints overlap.
To find this region, we need to draw each inequality on the coordinate plane.

• \( x = 0 \) is a vertical line.
• \( y = 0 \) is a horizontal line.
• \( x + y = 2 \) is a line with slope -1.
• \( 2x + 3y = 12 \) is a line with slope \(-\frac{2}{3}\).
• \( 3x + y = 12 \) is a line with slope -3.

After graphing, we shade the areas that satisfy each inequality.
The overlapping shaded area is the feasible region.
This region represents all possible solutions that satisfy the constraints.
It's essential to clearly identify this area because the optimal solution will lie within this region.

Constraints

Constraints are the inequalities or equations that limit the values of the variables.
In this exercise, the constraints are:

• \( x \geq 0 \)
• \( y \geq 0 \)
• \( x + y \geq 2 \)
• \( 2x + 3y \leq 12 \)
• \( 3x + y \leq 12 \)

Each constraint reduces the set of potential solutions.
They represent real-world limits, like resource availability or time restrictions.
By solving and plotting these constraints, we form the edges of the feasible region.
Ensuring all constraints are met is crucial for a viable solution.

Corner Points

Corner points, or vertices, are the points where the boundary lines of the feasible region intersect.
These points are crucial because, in linear programming, the optimal solution always lies at one of these vertices.
To find these points, solve the equations of the intersecting lines.
In this problem, the corner points are found by solving equations such as:

• \( x + y = 2 \) and \( x = 0 \) resulting in \((0, 2)\)
• \( x + y = 2 \) and \( y = 0 \) resulting in \( (2, 0) \)
• \( 3x + y = 12 \) and \( y = 0 \) resulting in \( (3, 0) \)
• \( 2x + 3y = 12 \) and \( x = 0 \) resulting in \( (0, 4) \)
• \( 2x + 3y = 12 \) and \( 3x + y = 12 \) resulting in \( (3, 3) \)

Evaluating the objective function at each corner point helps identify the optimal solution.

Optimization

Optimization in linear programming aims to find the best possible value of the objective function within the feasible region.
After identifying the corner points, we evaluate the objective function at each point.
In this exercise, we substitute each corner point into \( z = 5x + 4y \):

• At \((0, 2)\): \( z = 8 \)
• At \((2, 0)\): \( z = 10 \)
• At \((3, 0)\): \( z = 15 \)
• At \((0, 4)\): \( z = 16 \)
• At \((3, 3)\): \( z = 27 \)

The minimum value of \( z \) is 8 at \((0, 2)\).
Therefore, the optimal solution is \( x = 0 \) and \( y = 2 \).
Optimization helps in making efficient decisions within given limitations.