Inequalities and Linear Programming

Inequalities and Linear Programming

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

• An inequality compares two expressions using <, ≤, >, ≥. Multiplying or dividing both sides by a negative number reverses the sign; adding or subtracting keeps it the same.
• A linear inequality in two variables has the form ax + by < c (or ≤, >, ≥) and its solution set is a half-plane cut off by the boundary line ax + by = c.
• A Linear Programming Problem (LPP) asks for the maximum or minimum of a linear objective function Z = ax + by subject to linear constraints and x ≥ 0, y ≥ 0.
• The solution set of all constraints is the feasible region; the corner-point theorem says the optimum of Z occurs at a vertex (corner point) of this region.
• NECO SSCE tip: NECO almost always frames this as a word problem (diet, production, transport) requiring you to draw the feasible region, read its corners from the graph, then evaluate Z at each.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Linear Inequalities on the Plane

A linear inequality in x and y represents one side of a straight line. The line ax + by = c is the boundary line; everything on one side satisfies the inequality. To decide which side, test a simple point — usually (0, 0) — into the inequality:

• If (0, 0) satisfies ax + by < c, shade the side containing the origin.
• If it does not, shade the opposite side.

A strict inequality (< or >) gives an open boundary drawn as a dashed line — points on the line are excluded. A weak inequality (≤ or ≥) gives a closed boundary drawn as a solid line.

Solving One-Variable Inequalities Algebraically

Treat the sign as an equals sign to isolate the variable, then check the direction by substituting one value. The single most-tested rule: dividing both sides by a negative number flips the inequality.

Systems and the Feasible Region

A system of two or more linear inequalities is solved by drawing each half-plane and shading their intersection. The result is the feasible region — it may be:

• a bounded polygon (triangle, quadrilateral, etc.),
• an unbounded region (extending to infinity), or
• empty (no feasible region, hence no solution).

The Linear Programming Problem

An LPP has three ingredients:

1. Decision variables x, y (almost always x ≥ 0, y ≥ 0).
2. A linear objective function Z = ax + by to maximise or minimise.
3. Linear constraints such as 2x + y ≤ 8 and x + 3y ≤ 9.

The corner-point method (also called the vertex method) evaluates Z at every corner of the feasible region and picks the largest (for maximum) or smallest (for minimum) value.

Exam Pattern at NECO

NECO SSCE Mathematics (Paper II, ~3% weight) typically gives one 10–15 mark question: a word problem (factory output, blending, transport) that you convert into an LPP, sketch the feasible region, list the vertices, and state the optimal value. Always report both the optimal Z and the values of x, y that achieve it.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Edge Cases in Graphing

• Parallel constraints (e.g. x + y ≤ 5 and x + y ≥ 2) produce a strip; the feasible region is the part of the strip also satisfying the other constraints.
• Redundant constraints never touch the feasible region — they can be ignored but should still be written in your final LPP formulation.
• An unbounded feasible region can still yield a finite maximum (if Z increases towards the feasible interior) or no finite maximum at all (if Z grows without bound in the feasible direction). For a minimum on an unbounded region, the optimum exists only if Z has a lower bound within the feasible set.
• When a constraint line passes through the origin (ax + by ≤ 0), the test point (0,0) gives equality, so use (1, 0) or (0, 1) instead.

Worked Mini-Example

Maximise Z = 3x + 2y subject to x + y ≤ 6, 2x + y ≤ 8, x ≥ 0, y ≥ 0.

Solving the boundaries pairwise gives vertices (0, 0), (4, 0), (2, 4), (0, 6). Evaluating Z:

Vertex	Z = 3x + 2y
(0, 0)	0
(4, 0)	12
(2, 4)	14 ← maximum
(0, 6)	12

Maximum is Z = 14 at (x, y) = (2, 4).

Common Mistakes

• Forgetting to reverse the sign when multiplying through by -1 (e.g. -2x < 6 becoming x > -3, not x < -3).
• Shading the wrong half-plane because the test point was on the boundary itself.
• Reading vertices off a graph that is too small — NECO accepts graphs, but a vertex estimated between gridlines loses marks.
• Reporting only the optimal Z without the x, y values, or vice-versa — NECO marks require both.

Practice Prompts

1. A bakery makes cakes (x) and bread (y). Profit is ₦500 per cake and ₦300 per loaf. Constraints: 2x + y ≤ 10 (flour), x + 2y ≤ 8 (oven hours), x, y ≥ 0. Find the maximum profit and the production mix.
2. Minimise Z = 4x + 3y subject to x + y ≥ 4, 2x + y ≥ 6, x ≥ 0, y ≥ 0. State whether the feasible region is bounded and find the minimum.

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