HKDSE Maths | Inequalities and Linear Programming Prep Guide (for tutors) | Feasible Region, Vertex Method, Maximum Profit

2026-11-01

Tutor lesson-prep guides · Secondary Maths

Students can draw a straight line, but the moment a “less than or equal” sign appears they shade the wrong side, and in word problems they swap “at most” with “at least”. This lesson does three things: turn each inequality into a boundary with a direction, draw the feasible region, and hand the answer only to the vertices.

Objectives — Students can: (1) draw the boundary line for an inequality and decide which side to shade; (2) draw the feasible region and label its vertices; (3) use the vertex method to maximise or minimise P = ax + by; (4) translate a word problem into three constraints plus an objective function.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Equation of a straight line	What m means in y = mx + c	the gradient
Solving simultaneous equations	How to find where two lines meet	substitution or elimination
Direction of an inequality	What happens when you multiply by a negative	the sign flips

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Inequalities as graphs	Draw the line first, then test a point	substitute (0,0)
10-24 min	Feasible region	Stack the three constraints	vertex (6, 4)
24-40 min	Vertex method	Test vertices only	maximum P = 26
40-52 min	Word problems	One sentence, one constraint	“at most” means <=, "at least" means >=
52-60 min	Practice and exit check	Three questions unaided	one wrong means redo act two

03 Teaching script (read it out)

Scene 1: Act one: an inequality is a boundary with a direction

Treat the inequality as an equation, draw the line, then test a point (usually the origin) to decide the shading. For x + y <= 10, the line is x + y = 10; substituting (0,0) gives 0 <= 10, which is true, so shade the side containing the origin. If the inequality is strict, such as y > 2x, use a dashed line because points on the line do not satisfy it.
Ask the student: why test a point instead of reading the sign? (Because the coefficients of x and y can be negative, so the sign alone misleads.) Student task: draw 2x + y <= 16 and write down two integer points on the boundary.

Scene 2: Act two: the feasible region is the overlap of half-planes

All three constraints must hold at once, so the feasible region is the overlap of three half-planes. With x + y <= 10, 2x + y <= 16 and x >= 2 the region is a quadrilateral with vertices (2, 0), (8, 0), (6, 4) and (2, 8). To find a vertex, solve two boundary lines simultaneously, or pair one boundary with x = constant or y = 0.
Ask the student: why must the feasible region be convex? (It is an intersection of half-planes bounded by straight lines.) Student task: list the four vertices and check each one against all three inequalities.

Scene 3: Act three: the optimum sits on a vertex

P = 3x + 2y takes the values 6, 24, 26 and 22 at the four vertices. The maximum is 26 at (6, 4) and the minimum is 6. A linear function on a convex polygon always attains its extremes at vertices, so interior points never need testing, but every vertex must be checked.
Ask the student: what does it mean when two vertices give the same value? (The objective is parallel to that edge, so the whole edge is optimal.) Student task: change the objective to 2x + 3y and recompute the maximum.

04 Question types and worked steps

Type	Example	Steps	Answer
Graphing an inequality	given the inequality	draw boundary, test a point	decide solid or dashed first
Finding vertices	given three constraints	solve pairs simultaneously	(6, 4)
Vertex method	given the objective	substitute every vertex	maximum 26
Word-problem modelling	given a scenario and limits	one inequality per sentence	“at least” becomes >=

Sanity check (teach this): The vertex (6, 4) gives 2x + y = 16, exactly on the boundary, and x + y = 10, also on it, so both are tight as expected. P = 3 x 6 + 2 x 4 = 26, larger than 24 and 22 at neighbouring vertices, so the direction is consistent. The feasible region must be bounded; if the constraints fail to enclose an area there is no finite maximum, which is a standard trap. For integer-answer questions, check whether vertices are integral; if not, test nearby integer points.

05 Independent practice (3 questions)

Question	Answer	What to watch
How many vertices does the region x + y <= 10, 2x + y <= 16, x >= 2 have	4	intersect boundary lines
Maximum of P = 3x + 2y	26	at (6, 4)
Maximum of P = 2x + 3y on the same region	26	substitute every vertex again

06 Common mistakes and what to say

Mistake	Student writes	Say this
Shading the wrong side	shading away from the origin for x + y <= 10	“Substitute (0,0): 0 <= 10 is true, so shade the side with the origin."
Using a solid line for a strict inequality	y > 2x drawn solid	“Strict inequalities exclude the boundary, so the line must be dashed.”
Testing only one vertex	submitting (8, 0) because it looks largest	“All four vertices must be evaluated before you can claim a maximum.”
Missing a constraint in a word problem	using only the capacity limit	“Every at least or at most sentence is one inequality; miss one and the region is wrong.”

07 Homework (paper, section and question numbers)

Secondary Maths chapter index — inequalities set, questions 2 and 4
Prep guide master index — other chapters in this subject
HKDSE Maths practice papers — linear programming, questions 1 to 3

08 Exit check (last 5 minutes)

Question	Answer	Decision
Intersection of x + y <= 10 and 2x + y <= 16	(6, 4)	three correct means move to word problems
Maximum of P = 3x + 2y	26	one wrong means reteach the vertex method
Number of vertices of the region	4	two or more wrong means return to act one

09 Tutor FAQ

Why not test interior points?

The objective is linear, so on a convex polygon its maximum and minimum always occur at vertices; interior points only give intermediate values.

What if the region is unbounded?

Without an enclosing boundary the objective may grow without limit, so such questions only ask for a minimum.

How is integer programming different?

It requires x and y to be whole numbers, so the best point may sit next to the optimal vertex and nearby integer combinations must be tested.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.