An inequality in two variables shades half the plane
The equation x + y = 10 is a line. The inequality x + y ≤ 10 is that line together with everything on one side of it — a half-plane.
To decide which side, take any convenient test point not on the line and check whether it satisfies the inequality. The origin is almost always the easiest, and it works whenever the line does not pass through it. If the test point works, shade its side; if not, shade the other.
Draw the boundary solid for ≤ or ≥, because those points are included, and dashed for < or >, because they are not.
Shade in or shade out?
Some textbooks shade the region that satisfies the inequality; others shade the region that fails it, leaving the answer white. Both are accepted, but the two conventions look identical on paper and opposite in meaning — so write a key on your diagram saying which you have used. With four or five constraints, shading the unwanted regions keeps the answer visible instead of burying it under five layers.
The feasible region
Corner point — A vertex of the feasible region, found where two boundary lines cross. Solve those two equations simultaneously to get its exact coordinates — reading them off a hand-drawn graph is not accurate enough for full marks.
A system of inequalities is satisfied only where all the half-planes overlap. That overlap is the feasible region, and every point in it is a possible solution to the problem.
Most real problems carry two constraints that are easy to forget: x ≥ 0 and y ≥ 0. You cannot make a negative number of chairs. Those two restrict everything to the first quadrant, and omitting them produces an unbounded region and a nonsensical answer.
The shaded polygon satisfies every constraint at once. Slide k and watch the objective line sweep across without changing its gradient — the last corner it touches on its way out is the optimum.
Why the optimum is always at a corner
The objective function is the quantity to be maximised or minimised — profit, cost, time. Setting it equal to a constant gives a straight line, and changing the constant slides that line across the plane without changing its gradient.
Push the line in the direction of increasing value and it eventually leaves the feasible region. The last point of contact is where the maximum occurs, and because the region is a polygon, that last point is a corner — unless the objective line happens to be parallel to an edge, in which case the whole edge is optimal and every point on it gives the same value.
This is why the method works: instead of testing infinitely many points, you evaluate the objective at each corner and take the best.
A complete worked problem
Word problems are the whole point of this chapter, and translating the English into inequalities is where most of the marks sit. Define your variables in writing before anything else.
A workshop makes tables and chairs. Each table needs 4 hours of carpentry and 2 hours of finishing; each chair needs 2 hours of carpentry and 1 hour of finishing. There are 40 carpentry hours and 18 finishing hours available. Profit is Rs 700 per table and Rs 300 per chair. How many of each should be made?
- Let x = number of tables and y = number of chairs.Write the definitions down. An unlabelled answer of "8 and 4" cannot be marked.
- Carpentry:
4x + 2y ≤ 40, so2x + y ≤ 20. Finishing:2x + y ≤ 18. Alsox ≥ 0,y ≥ 0.Simplify each constraint as you write it; smaller numbers make the corner calculations easier. - Notice
2x + y ≤ 18is stricter than2x + y ≤ 20, so the carpentry limit is never reached — it is a redundant constraint.Spotting a redundant constraint is worth a comment; it also means one fewer line to draw. - Corners of the region:
(0, 0),(9, 0)and(0, 18).2x + y = 18 meets the axes at x = 9 and y = 18. - Profit
P = 700x + 300y. At (0,0): 0. At (9,0): 6300. At (0,18): 5400.Evaluate at every corner without trying to guess which will win. - Maximum profit Rs 6300, by making 9 tables and no chairs.A corner answer of zero for one product is a perfectly legitimate result and often the correct one.
9 tables and 0 chairs, for a maximum profit of Rs 6300
Whole numbers are not guaranteed
The corner where two lines cross may have fractional coordinates, and you cannot make 4.5 chairs. When the answer must be a whole number, test the integer points nearest that corner inside the region — the true optimum may not be the corner itself. Say what you have done; the examiner is looking for the awareness, not just the number.
Before you leave this chapter
- Test the origin to decide which side of a boundary to shade; solid line for ≤ and ≥, dashed for < and >.
- Always include x ≥ 0 and y ≥ 0 in a practical problem.
- The feasible region is where every constraint holds at once.
- The optimum of a linear objective is at a corner — find each corner by solving two equations simultaneously.
- State the optimal value AND the point at which it occurs.