Question

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

Short answer

Maximum value of \(z\) is 20 at the point (0, 4).

Step-by-step solution

Identify the Objective Function

The objective is to maximize the function given by \( z = 3x + 5y \). This is the function we want to find the maximum value for, subject to the given constraints.

List the Constraints

The constraints provided are: 1. \( x \geq 0 \)2. \( y \geq 0 \)3. \( x + y \geq 2 \)4. \( 2x + 3y \leq 12 \)5. \( 3x + 2y \leq 12 \)

Graph the Constraints

Plot the inequalities on a graph to find the feasible region. This area will be bounded by the lines: \(x = 0\), \(y = 0\), \(x + y = 2\), \(2x + 3y = 12\), and \(3x + 2y = 12\). The feasible region is where all these inequalities overlap.

Find the Corner Points

Determine the points where these lines intersect to identify the vertices (corner points) of the feasible region. Solving the equations, we get the points: (0, 4), (0, 0), (2, 0), (3, 2), and (6/5, 12/5).

Evaluate the Objective Function

Plug the coordinates of each vertex into the objective function \( z = 3x + 5y \). For the points:1. (0, 4): \( z = 3(0) + 5(4) = 20 \)2. (0, 0): \( z = 3(0) + 5(0) = 0 \)3. (2, 0): \( z = 3(2) + 5(0) = 6 \)4. (3, 2): \( z = 3(3) + 5(2) = 19 \)5. (6/5, 12/5): \( z = 3(6/5) + 5(12/5) = 18 \)

Determine the Maximum Value

Compare the values obtained for each vertex. The maximum value of \( z \) is 20, which occurs at the point (0, 4).

Key concepts

Objective Function

In linear programming, we aim to optimize (maximize or minimize) a specific function, known as the objective function. An objective function is a mathematical expression that describes the problem's goal. In our exercise, the objective function is given as \( z = 3x + 5y \). This equation quantifies what we're trying to maximize — for instance, profit, time, or another measurable quantity. We need to find the values of \( x \) and \( y \) that make this function as large as possible, subject to certain restrictions.

Constraints

Constraints are the conditions or restrictions placed on the decision variables (in our case, \( x \) and \( y \)). These constraints define the limits within which the objective function must be optimized. For our problem, the constraints are as follows:

• \( x \geq 0 \): \( x \) must be non-negative.


• \( y \geq 0 \): \( y \) must be non-negative.


• \( x + y \geq 2 \): The sum of \( x \) and \( y \) must be at least 2.


• \( 2x + 3y \leq 12 \): A combined constraint representing a linear relationship between \( x \) and \( y \).


• \( 3x + 2y \leq 12 \): Another linear relationship restricting \( x \) and \( y \).

These inequalities need to be satisfied simultaneously when solving the problem.

Feasible Region

The feasible region is the set of all possible points that satisfy the constraints. To identify this region, we graph each inequality and look for the area where all these constraints overlap. This area will be bounded by the lines \( x = 0 \), \( y = 0 \), \( x + y = 2 \), \( 2x + 3y = 12 \), and \( 3x + 2y = 12 \).

Only points inside (or on the boundary of) this feasible region are potential solutions to the problem. Any point outside of this area would violate at least one of the constraints.

In our exercise, we plot each constraint to find where they intersect and shape our feasible region.

Corner Points Evaluation

When dealing with linear programming, the maximum or minimum value of the objective function will always lie at one of the vertices (corner points) of the feasible region. To find these points, we solve the equations where the constraint lines intersect. For our problem, the intersection points (corner points) are:

• \( (0, 4) \)


• \( (0, 0) \)


• \( (2, 0) \)


• \( (3, 2) \)


• \( (6/5, 12/5) \)

We then take these points and evaluate the objective function at each one.

For example, evaluating at point \( (0, 4) \), we get\( z = 3(0) + 5(4) = 20 \). Doing this for all points, we find that the maximum value is 20, which occurs at \( (0, 4) \). This point is our optimal solution.