Explore: Graphing for Linear Programming Solutions

If a linear programming problem does not provide a graph, you must create the graph yourself using the provided system of inequalities. 
Once the graph is created, label the vertices so you can find the minimum and/or maximum using the objective function. 
Use substitution to confirm that the vertices are correct. To confirm, replace the variables with the point (or vertex) where the two equations or inequalities intersect. 
Graphs of systems of inequalities can have feasibility regions that are bounded or unbounded .

A bounded region is an enclosed , shaded region with vertices that provide at least one minimum and one maximum using the objective function.

An unbounded region continues infinitely in at least one direction with vertices that provide a minimum or a maximum, but not both. The minimum or maximum depends on the direction of the shaded region.

Find the minimum and maximum values using the objective function: $f (x, y) = 5 y - 2 x .$ Explain if the system is bounded or unbounded.

$y \leq - 2 x + 6 x + y \leq 4 x \geq 0 y \geq 0$

Plan
Graph system of inequalities.

Name all the vertices.

Evaluate the objective function using the vertices.

Name the minimum and maximum values.

Implement

$f (0) = 5 (0) - 2 (0) f (0, 0) = 0$

$f (0, 4) = 5 (4) - 2 (0) f (0, 4) = 20 maximum$

$f (2, 2) = 5 (2) - 2 (2) f (2, 2) = 6$

$f (3, 0) = 5 (0) - 2 (3) f (3, 0) = - 6 minimum$

Explain

This is a bounded system. The maximum is (0, 4) and the minimum is (3, 0) .

Find the minimum and maximum values using the objective function: $h (x, y) = x + y - 1 .$

$y \leq 2 x - 3 y \geq - 3 y \geq \frac{1}{3} x - 4$

$h (x, y) = x + y - 1 h (0, - 3) = (0) + (- 3) - 1 h (0, - 3) = - 4 \begin{matrix} h (3, - 3) = (3) + (- 3) - 1 h (3, - 3) = - 1 \end{matrix}$

The minimum is (0, –3). There is no maximum because the system is unbounded.