CUET UG 2025 — Mathematics Algebra

The corner points (vertices) of the feasible region are found where two constraint lines intersect or where a constraint line meets an axis.

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Given constraints:

$2x - y \leq -5$2x−y≤−5 which gives $y \geq 2x + 5$y≥2x+5

$3x + y \geq 3$3x+y≥3 which gives $y \geq 3 - 3x$y≥3−3x

$2x - 3y \leq 12$2x−3y≤12 which gives $y \geq \frac{2x - 12}{3}$y≥32x−12​

$$x \geq 0, y \geq 0$$x≥0,y≥0

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For points on the y-axis, set $x = 0$x=0:

From $y \geq 2(0) + 5$y≥2(0)+5:

$$y \geq 5$$y≥5

Boundary point: $(0, 5)$(0,5)

From $y \geq 3 - 3(0)$y≥3−3(0):

$$y \geq 3$$y≥3

Boundary point: $(0, 3)$(0,3)

From $y \geq \frac{2(0) - 12}{3}$y≥32(0)−12​:

$$y \geq -4$$y≥−4

This does not restrict $y$y further in the first quadrant.

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For points on the x-axis, set $y = 0$y=0:

From $2x - 0 \leq -5$2x−0≤−5:

$$x \leq -2.5$$x≤−2.5

This is not possible since $x \geq 0$x≥0.

From $3x + 0 \geq 3$3x+0≥3:

$$x \geq 1$$x≥1

Boundary point: $(1, 0)$(1,0)

From $2x - 3(0) \leq 12$2x−3(0)≤12:

$$x \leq 6$$x≤6

Boundary point: $(6, 0)$(6,0)

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The corner points of the feasible region are:

$$(0, 3)$$(0,3)

$$(0, 5)$$(0,5)

$$(1, 0)$$(1,0)

$$(6, 0)$$(6,0)