Linear inequalities and programming

Boundary, then test point

Replace an inequality with an equation to draw its boundary. Use a solid boundary for ≤ or ≥, and a dashed boundary for < or >. Choose a test point away from the line and substitute it into the original inequality. Shade the side containing the point if the statement is true.

For 2x + y ≤ 8, the boundary passes through (4, 0) and (0, 8). The origin gives 0 ≤ 8, so shade its side and include the line. For x > 2, draw a dashed vertical line and shade to the right. Dividing by a negative reverses an inequality: −2y ≤ 6 − 2x becomes y ≥ x − 3.

Find the overlap and its corners

For a system, keep only points that satisfy every inequality. Original example: x ≥ 0, y ≥ 0, x ≤ 4, y ≤ 3, x + y ≤ 6.

The feasible polygon has corners (0, 0), (4, 0), (4, 2), (3, 3), and (0, 3). For instance, x = 4 and x + y = 6 meet at (4, 2). The candidate (4, 3) fails x + y ≤ 6, so it is not a feasible corner.

Find boundary intersections, then test each candidate against all the constraints. If decimal coefficients are used, keep extra precision during solving and round only the final coordinates as requested.

Evaluate an objective at the feasible corners

Original example: maximize P = 5x + 4y over the polygon above. The five corner values, in the order listed, are 0, 20, 28, 27, and 12. The maximum is 28 at (4, 2). For a minimum over this same polygon, compare the same values in the opposite direction.

For a nonempty closed bounded polygon, a linear objective attains an optimum at a corner. If adjacent optimal corners tie, all points on the joining edge also tie. The objective does not replace the constraints.

Build a model from a small application

Original model: a workshop makes x small kits and y large kits. Each small kit uses 2 packing hours; each large kit uses 3. At most 18 hours are available, at most 5 small kits may be made, and profit is $7 per small kit and $10 per large kit.

Define variables and units before writing coefficients. Resource limits become constraints; per-unit profit belongs in the objective. This is a continuous graphical model. If a problem requires whole-number quantities, a fractional recommendation needs an additional check against that requirement.

Empty and unbounded mean different things

The constraints x ≥ 4 and x ≤ 1 cannot both hold, so the feasible region is empty. No feasible solution exists.

The region x ≥ 0, y ≥ 0 is unbounded. Maximizing x + y there has no finite maximum because the value can grow without limit. Minimizing x + y over that same region has minimum 0 at (0, 0). An unbounded region alone does not prove the requested objective has no optimum.

Try it before checking

Use the Practice quiz to read a graph, test points, choose a model, and compare objective values. Then use Explain-it for short written reasoning without answer choices. These are original practice problems, not answers to the assigned homework.