MHT-CET Maths · Linear Programming

Reading Constraints Off a Shaded Region

Given a shaded region, recover its inequalities: write each boundary line from its intercepts, then pick a point inside the shading and test it against each line to fix ≥ or ≤.

All Linear Programming notes MHT-CET Maths strategy

Why this matters

11 PYQs at 18% HARD — the only page in the chapter with HARD questions, and both of them are here. The stems give a figure with three to five boundary lines and four option sets that differ only in the direction of one or two inequalities. The method is the half-plane test run backwards: one interior point, tested against one line at a time, eliminates the wrong options without ever sketching.

Concept 1 of 3: The Boundary Line From Its Intercepts: x/a + y/b = 1

A line cutting the axes at

(a, 0)

and

(0, b)

\dfrac{x}{a} + \dfrac{y}{b} = 1

, i.e.

bx + ay = ab

. Read the intercepts off the figure and the equation writes itself; a line through the origin with slope

m

y = mx

Definition

Intercepts $4$ and $3$ : $\dfrac{x}{4} + \dfrac{y}{3} = 1 \Rightarrow 3x + 4y = 12$ . Intercepts $8$ and $4$ : $x + 2y = 8$ . Intercepts $5$ and $2$ : $2x + 5y = 10$ .
A line through the origin and $(1, 1)$ is $y = x$ , written $y - x = 0$ or $x - y = 0$ in the options.
A line through $(-1, 0)$ and $(0, 1)$ is $y = x + 1$ , i.e. $y - x = 1$ ; the sign of the intercept decides the sign of the constant.
Horizontal and vertical boundaries are $y = 3$ and $x = 4$ ; these become $y \le 3$ or $y \ge 3$ in the constraint set.
Match each option's lines to the figure's lines FIRST; an option whose line does not appear in the figure at all is eliminated before any inequality is tested.

Intercept form

\frac{x}{a} + \frac{y}{b} = 1 \iff bx + ay = ab

Worked example

A boundary line in a figure crosses the

x

-axis at

6

and the

y

-axis at

2

. Write its equation.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q108Moderate

Example 1 · Linear Programming · Reading Constraints Off a Shaded Region

If feasible region is as shown in the figure, then the related inequalities are

Swapping the intercepts

3x + 4y = 12

has

x

-intercept

4

, not

3

. The coefficient of

x

is the

y

-intercept's number and vice versa; the option built on the swap is always offered.

Drill 2 more on the boundary line from its intercepts: x/a + y/b = 1

Concept 2 of 3: Fix ≥ or ≤ With One Point Inside the Shading

Pick a point visibly inside the shaded region, away from every boundary. Substitute it into each line's expression: if the value is below the constant the region is on the

\le

side, if above, the

\ge

side.

Definition

Region with boundaries $x + 2y = 50$ , $2x + y = 100$ , $2x - y = 0$ , interior point $(10, 40)$ : $10 + 80 = 90 \ge 50$ , $20 + 40 = 60 \le 100$ , $20 - 40 = -20 \le 0$ . So $x + 2y \ge 50$ , $2x + y \le 100$ , $2x - y \le 0$ .
If the origin is NOT inside the shading, the origin test still works in reverse: a line the shading is on the far side of gets the direction the origin FAILS.
Two constraints of the form $x - y \ge 0$ , $x + y \ge 0$ describe a wedge to the right of the origin between the lines $y = x$ and $y = -x$ ; $(2, 0)$ settles both signs at once.
Non-negativity: if the shading touches an axis, $x, y \ge 0$ is part of the answer; options that omit it or misprint it ( $xy \ge 0$ ) are the same option in disguise.

Direction test

P \text{ inside the shading:}\quad aP_x + bP_y < c \Rightarrow ax + by \le c,\qquad > c \Rightarrow ax + by \ge c

Worked example

A shaded region has boundaries

x + y = 6

x - y = 2

and the axes, and the point

(3, 1)

lies inside it. Write the constraints.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 2 · Q118Moderate

Example 2 · Linear Programming · Reading Constraints Off a Shaded Region

The shaded region in the following figure is the solution set of the inequations

Testing a point on the boundary

(10, 40)

is a good test point precisely because it is nowhere near a line. A point on a boundary gives equality and decides nothing.

Drill 2 more on fix ≥ or ≤ with one point inside the shading

Concept 3 of 3: Four or Five Lines: Eliminate Options One Line at a Time

With five boundaries and four options, do not reconstruct the whole region. Take the options' lines one at a time, decide the direction with your interior point, and strike out every option that disagrees — usually two lines are enough.

Definition

Boundaries $3x + 8y = 24$ , $4x + 5y = 20$ , $5x + 3y = 15$ with the shading above the first line and below the other two: a point inside, $(1, 2.7)$ , gives $3 + 21.6 = 24.6 \ge 24$ , $4 + 13.5 = 17.5 \le 20$ , $5 + 8.1 = 13.1 \le 15$ . So $3x + 8y \ge 24$ , $4x + 5y \le 20$ , $5x + 3y \le 15$ — option (D). The region is thin, so the test point must be read off the figure, not guessed; that is what makes these two stems the chapter's only HARD ones.\n- The reliable route: identify which side of EACH line the shading sits by looking at the line and the shading directly (above/below, left/right), then convert: above a line with positive coefficients is $\ge$ , below is $\le$ .
Six-constraint stem with $2x + 3y = 6$ , $3x + 6y = 18$ , $x - 3y = 3$ , $-x + 2y = 2$ : shading is above $2x + 3y = 6$ ( $\ge$ ), below $3x + 6y = 18$ ( $\le$ ), above $x - 3y = 3$ ( $x - 3y \le 3$ , because larger $y$ makes $x - 3y$ smaller), below $-x + 2y = 2$ ( $\le$ ).
Lines with a NEGATIVE coefficient reverse the intuition: 'above' the line $x - 3y = 3$ means $x - 3y \le 3$ . Compute the sign at a point rather than trusting above/below.

Elimination

\text{for each line: decide the side} \to \text{strike every option with the other sign} \to \text{stop when one option is left}

Worked example

Options (A)–(D) share the lines

x + y = 5

x - y = 1

y = 3

and differ in signs. The shading lies below

x + y = 5

, above

x - y = 1

, and below

y = 3

. Write the constraints.

Practise this conceptself-check

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 1 · Q129Hard

Example 3 · Linear Programming · Reading Constraints Off a Shaded Region

The function to be maximized is given by

Z=3x+2y

. The feasible region for this function is the shaded region given below, then the linear constraints for this region are given by

'Above the line' read as ≥ when a coefficient is negative

Above

x - 3y = 3

x - 3y \le 3

. Evaluate the expression at a point; never convert above/below to a sign by reflex.

Drill 2 more on four or five lines: eliminate options one line at a time

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

The Boundary Line From Its Intercepts: x/a + y/b = 1
Intercept form
$\frac{x}{a} + \frac{y}{b} = 1 \iff bx + ay = ab$
Fix ≥ or ≤ With One Point Inside the Shading
Direction test
$P \text{ inside the shading:}\quad aP_x + bP_y < c \Rightarrow ax + by \le c,\qquad > c \Rightarrow ax + by \ge c$
Four or Five Lines: Eliminate Options One Line at a Time
Elimination
$\text{for each line: decide the side} \to \text{strike every option with the other sign} \to \text{stop when one option is left}$

Watch out for (3)

Swapping the intercepts
→ The Boundary Line From Its Intercepts: x/a + y/b = 1
Testing a point on the boundary
→ Fix ≥ or ≤ With One Point Inside the Shading
'Above the line' read as ≥ when a coefficient is negative
→ Four or Five Lines: Eliminate Options One Line at a Time

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