CUET UG 2025 — Mathematics Algebra

The constraints are:

$$x + y \leq 6$$x+y≤6

$$x \geq 0$$x≥0

$$y \geq 0$$y≥0

These create a triangular feasible region.

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The corner points of the feasible region are found where the constraint lines intersect:

At $x = 0$x=0 and $y = 0$y=0: $(0, 0)$(0,0)

At $y = 0$y=0 and $x + y = 6$x+y=6:

$$x + 0 = 6$$x+0=6

$$x = 6$$x=6

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

At $x = 0$x=0 and $x + y = 6$x+y=6:

$$0 + y = 6$$0+y=6

$$y = 6$$y=6

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

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Evaluating $z = 3x + 4y$z=3x+4y at each corner point:

At $(0, 0)$(0,0):

$$z = 3(0) + 4(0)$$z=3(0)+4(0)

$$z = 0$$z=0

At $(6, 0)$(6,0):

$$z = 3(6) + 4(0)$$z=3(6)+4(0)

$$z = 18$$z=18

At $(0, 6)$(0,6):

$$z = 3(0) + 4(6)$$z=3(0)+4(6)

$$z = 24$$z=24

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The maximum value of $z$z is $24$24, occurring at the point $(0, 6)$(0,6).