CUET UG 2025 — Mathematics Algebra

The feasible region is in the first quadrant where both x and y are positive.

This gives: $x \geq 0, y \geq 0$x≥0,y≥0

This eliminates Option 1 (which has $x \leq 0, y \leq 0$x≤0,y≤0).

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The feasible region is bounded by two lines:

Line 1: $3x + 4y = 24$3x+4y=24

Line 2: $x + 2y = 10$x+2y=10

Finding where the lines meet the axes:

For $3x + 4y = 24$3x+4y=24:

When $x = 0$x=0: $4y = 24$4y=24, so $y = 6$y=6

When $y = 0$y=0: $3x = 24$3x=24, so $x = 8$x=8

For $x + 2y = 10$x+2y=10:

When $x = 0$x=0: $2y = 10$2y=10, so $y = 5$y=5

When $y = 0$y=0: $x = 10$x=10

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Using the origin $(0,0)$(0,0) as a test point to determine the inequalities:

For line $3x + 4y = 24$3x+4y=24:

At origin: $3(0) + 4(0) = 0$3(0)+4(0)=0

Since $0 < 24$0<24 and the origin is inside the shaded region relative to this line:

$$3x + 4y \leq 24$$3x+4y≤24

For line $x + 2y = 10$x+2y=10:

At origin: $0 + 2(0) = 0$0+2(0)=0

Since $0 < 10$0<10 but the origin is outside the shaded region relative to this line:

$$x + 2y \geq 10$$x+2y≥10

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The complete set of constraints:

$$3x + 4y \leq 24$$3x+4y≤24

$$x + 2y \geq 10$$x+2y≥10

$$x \geq 0, y \geq 0$$x≥0,y≥0

This matches Option 4.