Build a feasible region from constraints

Try

A fictional workshop makes x small kits and y large kits. Each small kit uses two units of material and each large kit one. There are eight units available. The inequality 2x+y≤8 says that total use cannot exceed the supply. It does not require using every unit.

Test (2,3): the use is 7, so it passes. Test (4,1): the use is 9, so it fails. A shaded region should encode these decisions, not replace them. Assume first that x and y are real numbers so that we can understand the geometry; actual whole-kit decisions will need an additional integer condition.

Predict

The boundary 2x+y=8 meets the axes at (0,8) and (4,0). The origin gives 0≤8, so the half-plane containing the origin is included. Because equality is allowed, draw a solid boundary. For 2x+y<8 the same side is selected but the boundary is dashed and excluded.

Why does one test point identify a side? Rewriting gives y≤8−2x. At a fixed x, every point below the line has a smaller y and hence satisfies the inequality. Above it, the reverse holds. For a vertical boundary, such as x≥2, compare x directly; there is no need to force it into a slope formula.

Experiment

Use the live construction to change one relation or bound. Its playground constraints differ from the fixed workshop example. Predict whether the probe remains feasible, then inspect each substitution check. Compare a boundary point under a strict sign and an inclusive sign.

Construct the common feasible region

Each inequality selects a half-plane. Predict which side of each boundary contains the probe point, and whether the point satisfies all three. A strict inequality excludes its dashed boundary; an inclusive inequality includes its solid boundary.

Starting constraintsA bounded triangleAn unbounded regionConflicting constraintsA single feasible pointLoad constraintsReset region

Constraint 1

xRelation≤<≥>Bound-4-3-2-101234

Constraint 2

yRelation≤<≥>Bound-4-3-2-101234

Constraint 3

x + (1)yRelation≤<≥>Bound-4-3-2-101234

Probe x-6-5-4-3-2-10123456Probe y-6-5-4-3-2-10123456Reveal region and checks

Test the same point in every constraint

Probe: (3, 2). Predict before revealing.

Substitution checks for the probe

Constraint	Calculation	Holds?
1	Hidden	Hidden
2	Hidden	Hidden
3	Hidden	Hidden

• Boundary 1: x = 0; solid and included by this constraint. Other constraints may exclude parts of that line.
• Boundary 2: y = 0; solid and included by this constraint. Other constraints may exclude parts of that line.
• Boundary 3: x + (1)y = 4; solid and included by this constraint. Other constraints may exclude parts of that line.

The axes never rescale. A clipped edge of the displayed square is not an extra inequality. Use the constraints to decide global emptiness or boundedness; the picture alone cannot establish either. Integer dots represent discrete choices, while shading represents continuous choices. Temporary controls do not award completion.

Record one accepted point, one rejected point and the precise failed inequality. The displayed window is finite. If shading reaches its edge, that edge is not automatically a constraint or evidence of a largest possible value. The integer lattice helps inspect whole-number decisions; the entire shaded real region remains a different mathematical object.

Understand

Combine the workshop condition with x≥0, y≥0 and x+y<6. The feasible set is the intersection of all four conditions: a point must satisfy every one. It is not the union of individually allowed regions. The two oblique boundary lines meet where 2x+y=8 and x+y=6, giving x=2, y=4. That crossing is excluded because the second inequality is strict.

Candidate	Material use 2x+y	Kit total x+y	Result
(2,3)	7	5	Included
(2,4)	8	6	Excluded by the strict total limit
(4,0)	8	4	Included
(0,6)	6	6	Excluded by the strict total limit

These checks illustrate the region; the inequalities define all its points. Boundary intersections alone are not a sufficient description when some edges are open.

Three outcomes must be kept separate. The pair x+y≤1 and x+y>1 has no solution because the same number cannot meet both comparisons. The pair x≥0 and y≥0 is nonempty and unbounded: (t,0) remains allowed for arbitrarily large t. Adding x+y≤5 produces a bounded triangular region. “No closed polygon” does not mean “no solution.”

Apply

If kits cannot be divided, add that x and y are nonnegative integers. The real point (1.5,2) can lie in the shaded region but is not an admissible whole-kit decision. Do not round a continuous candidate without checking it again: rounding upward can violate a resource limit, and rounding downward can violate a minimum requirement.

For your own diagram, label each boundary with its equation and strictness, choose a test point away from it, shade the common region, and check a point on each relevant edge. Practice includes a fresh materials-and-time problem. Retain units beside the resource calculations: sheets and minutes are different constraints, not numbers to add together.

A false proof says that every point in a bounded screen window is below a maximum shown at the right edge. Repair it with a feasible family or a genuine mathematical bound. For x≥1 and y≥x, the points (t,t) remain feasible for every t≥1, regardless of the screen size.

Continue

Feasible regions describe what is allowed. Choosing a best feasible point requires an additional objective, such as minimizing cost or maximizing completed kits. The objective must not silently replace the original constraints.

You can proceed to graphical systems without completing an optimization course. A system of equations asks which boundaries meet exactly; a system of inequalities asks which sides overlap. Both require verification in the original statements, and both can be empty, have finitely many points, or contain infinitely many points.