CUET UG 2025 — Mathematics Calculus

The constraints are:

$$x \leq 6$$x≤6

$$y \leq 2$$y≤2

$x - y \leq 0$x−y≤0 (which means $x \leq y$x≤y)

$$x \geq 0$$x≥0

$$y \geq 0$$y≥0

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In Linear Programming, the maximum or minimum value occurs at a corner point of the feasible region.

The corner points are found by checking where the boundary lines intersect:

Origin: $(0, 0)$(0,0)

Where $x = 0$x=0 meets $y = 2$y=2: $(0, 2)$(0,2)

Where $x = y$x=y meets $y = 2$y=2:

If $y = 2$y=2 and $x = y$x=y, then $x = 2$x=2

Point: $(2, 2)$(2,2)

Note: The constraint $x \leq 6$x≤6 is not active since $x \leq y$x≤y and $y \leq 2$y≤2 means $x$x cannot exceed $2$2.

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Evaluating $Z = 2x + y$Z=2x+y at each corner point:

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

$$Z = 2(0) + 0 = 0$$Z=2(0)+0=0

At $(0, 2)$(0,2):

$$Z = 2(0) + 2 = 2$$Z=2(0)+2=2

At $(2, 2)$(2,2):

$$Z = 2(2) + 2$$Z=2(2)+2

$$Z = 4 + 2$$Z=4+2

$$Z = 6$$Z=6

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Comparing all values: $0$0, $2$2, and $6$6

The maximum value is $6$6 at point $(2, 2)$(2,2).