An equation like $x + y = 5$ is satisfied by the points on a line. An inequality like $x + y \leqslant 5$ is satisfied by every point on one side of that line — a whole region of the plane.
Dashed line for $<$ or $>$ — the boundary is not included
- Replace the inequality sign with $=$ to get the boundary line.
- Plot the line using the two axis intercepts (set $x=0$, then $y=0$).
- Draw it solid for $\leqslant$ / $\geqslant$, dashed for $<$ / $>$.
- Pick a test point not on the line — the origin if possible.
- If the test point satisfies the inequality, its side is the region. If not, the region is the other side.
Illustrate $24x + 28y \leqslant 400$ on a graph, for $x \geqslant 0$ and $y \geqslant 0$.
Illustrate $y > 2x$.
| Inequality | Boundary | Region (when it works out this way) |
|---|---|---|
| $x \geqslant 3$ | Vertical line $x = 3$ | To the right, solid |
| $y < 4$ | Horizontal line $y = 4$ | Below, dashed |
| $y \leqslant 2x + 1$ | Line of gradient $2$ | Below, solid |
| $x + y > 6$ | Line through $(6,0)$, $(0,6)$ | Away from the origin, dashed |
On one diagram, show the region satisfying $x \geqslant 0$, $y \geqslant 0$, $x + y \leqslant 8$ and $y \leqslant 2x$.
| The words say | You write |
|---|---|
| at most / no more than / maximum | $\leqslant$ |
| at least / no fewer than / minimum | $\geqslant$ |
| fewer than / under | $<$ |
| more than / over / exceeds | $>$ |
| cannot be negative | $\geqslant 0$ |
| at least twice as many $y$ as $x$ | $y \geqslant 2x$ |
A baker makes $x$ loaves and $y$ cakes. A loaf needs $400$ g of flour, a cake $250$ g, and only $30$ kg of flour is available. Write this as an inequality in its simplest form.
Boundary
Replace the sign with $=$.
Solid line
For $\leqslant$ and $\geqslant$.
Dashed line
For $<$ and $>$.
Intercepts
Set $x=0$, then $y=0$, to plot fast.
Test point
Use the origin unless the line passes through it.
Simplify first
Divide by a positive common factor.
Several at once
The answer is the overlap.
Vertices
Solve the boundary lines in pairs.
Label it
Mark the region "R" so the examiner knows which side you mean.
Hidden constraints
$x \geqslant 0$, $y \geqslant 0$ for physical counts.
Describe how you would draw the boundary for $y < 3x - 2$, and state which side is the region.
▶ Show solution
Draw $y = 3x - 2$ as a dashed line (the inequality is strict), through $(0,-2)$ with gradient $3$.
Test the origin: is $0 < -2$? No. So the origin is not in the region.
The region is the other side — below the line.
Find the $x$- and $y$-intercepts of the boundary of $5x + 3y \leqslant 45$.
▶ Show solution
$y = 0$: $5x = 45$, so $x = 9$ — the point $(9, 0)$.
$x = 0$: $3y = 45$, so $y = 15$ — the point $(0, 15)$.
Does the point $(4, 3)$ satisfy $2x + 5y \leqslant 22$?
▶ Show solution
$2(4) + 5(3) = 8 + 15 = 23$
$23 \leqslant 22$ is false, so no — the point lies outside the region.
Simplify the inequality $18x + 24y \leqslant 360$.
▶ Show solution
The highest common factor of $18$, $24$ and $360$ is $6$.
$3x + 4y \leqslant 60$
Dividing by a positive number, so the sign does not change.
Write inequalities for: "$x$ and $y$ are non-negative, their total is at most $20$, and there are at least three times as many $x$ as $y$."
▶ Show solution
$x \geqslant 0$, $y \geqslant 0$
$x + y \leqslant 20$
$x \geqslant 3y$
Careful: "three times as many $x$ as $y$" means $x$ is the bigger one, so $x \geqslant 3y$ — not $y \geqslant 3x$.
Find the point where the boundaries of $x + y \leqslant 10$ and $2x + y \leqslant 14$ intersect.
▶ Show solution
Solve $x + y = 10$ and $2x + y = 14$ simultaneously.
Subtracting: $x = 4$.
Then $y = 6$. The intersection is $(4, 6)$.
The region $R$ is defined by $x \geqslant 1$, $y \geqslant 1$ and $x + y \leqslant 5$. Find its three vertices.
▶ Show solution
$x=1$ with $y=1$: $(1, 1)$
$x=1$ with $x+y=5$: $(1, 4)$
$y=1$ with $x+y=5$: $(4, 1)$
Explain why the region defined by $x + y \leqslant 4$ and $x + y \geqslant 7$ is empty.
▶ Show solution
The first requires $x + y$ to be at most $4$; the second requires it to be at least $7$.
No number can be both $\leqslant 4$ and $\geqslant 7$, so no point satisfies both.
Geometrically, the two boundary lines are parallel (both have gradient $-1$) and the required half-planes point in opposite directions, so they never overlap.
A student writes the region $y \geqslant x^2$. Explain why this is not covered by AL9, and describe the region anyway.
▶ Show solution
AL9 concerns linear inequalities, whose boundaries are straight lines. $y \geqslant x^2$ has a curved boundary, so it is not a linear inequality.
The region is everything on or above the parabola $y = x^2$, with the parabola itself included (solid curve).
Check with a test point: $(0, 1)$ gives $1 \geqslant 0$ ✓, and $(0,1)$ is above the vertex, confirming the region is the inside of the parabola.
A workshop makes $x$ chairs and $y$ tables. Each chair takes $2$ hours of carpentry and $1$ hour of finishing; each table takes $5$ hours of carpentry and $1$ hour of finishing. There are $40$ carpentry hours and $12$ finishing hours available each week.
(a) Write down all four inequalities. (b) Find the vertices of the feasible region. (c) Explain why only whole-number points inside the region are actually usable. (d) Find the largest number of tables that can be made in a week.
▶ Show solution
(a) Carpentry: $2x + 5y \leqslant 40$
Finishing: $x + y \leqslant 12$
Non-negativity: $x \geqslant 0$, $y \geqslant 0$
(b) The vertices are where pairs of boundaries meet.
$(0,0)$ — both axes.
$y = 0$ with $x + y = 12$: $(12, 0)$. Check carpentry: $2(12) = 24 \leqslant 40$ ✓ so this is a genuine vertex.
$x = 0$ with $2x + 5y = 40$: $(0, 8)$. Check finishing: $8 \leqslant 12$ ✓ genuine.
The two slanted lines meet where $2x + 5y = 40$ and $x + y = 12$. From the second, $x = 12 - y$:
$2(12-y) + 5y = 40 \Rightarrow 24 + 3y = 40 \Rightarrow y = \tfrac{16}{3}$, and $x = \tfrac{20}{3}$.
Vertices: $(0,0)$, $(12,0)$, $\left(\tfrac{20}{3}, \tfrac{16}{3}\right)$, $(0,8)$.
(c) You cannot make a fraction of a chair or table, so only points with $x$ and $y$ whole numbers are usable. The vertex $\left(\tfrac{20}{3}, \tfrac{16}{3}\right)$ is not itself a possible production plan — it is a corner of the mathematical region only.
(d) Maximising $y$: the carpentry constraint gives $5y \leqslant 40$, so $y \leqslant 8$; the finishing constraint gives $y \leqslant 12$. The binding one is carpentry.
$y = 8$ is achievable, with $x = 0$: that uses $40$ carpentry hours and $8$ finishing hours, both within limits.
So at most $8$ tables per week.