CUET UG 2025 — Mathematics Algebra

The objective function is $z = 2x + 3y$z=2x+3y with constraints:

$$x - y \le 1$$x−y≤1

$$x + y \le 3$$x+y≤3

$$x \ge 0$$x≥0

$$y \ge 0$$y≥0

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The feasible region is bounded by the constraint lines. The corner points are found at the intersections of these boundary lines.

At the origin:

Point: $(0, 0)$(0,0)

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Where $x = 0$x=0 meets $x + y = 3$x+y=3:

$$0 + y = 3$$0+y=3

$$y = 3$$y=3

Point: $(0, 3)$(0,3)

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Where $y = 0$y=0 meets $x - y = 1$x−y=1:

$$x - 0 = 1$$x−0=1

$$x = 1$$x=1

Point: $(1, 0)$(1,0)

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Where $x - y = 1$x−y=1 meets $x + y = 3$x+y=3:

From $x - y = 1$x−y=1: $x = y + 1$x=y+1

Substituting into $x + y = 3$x+y=3:

$$(y + 1) + y = 3$$(y+1)+y=3

$$2y + 1 = 3$$2y+1=3

$$2y = 2$$2y=2

$$y = 1$$y=1

$$x = 2$$x=2

Point: $(2, 1)$(2,1)

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The corner points of the feasible region are: $(0, 0)$(0,0), $(0, 3)$(0,3), $(1, 0)$(1,0), and $(2, 1)$(2,1).

Evaluating the objective function at each corner point:

At $(0, 0)$(0,0): $z = 2(0) + 3(0) = 0$z=2(0)+3(0)=0

At $(0, 3)$(0,3): $z = 2(0) + 3(3) = 9$z=2(0)+3(3)=9

At $(1, 0)$(1,0): $z = 2(1) + 3(0) = 2$z=2(1)+3(0)=2

At $(2, 1)$(2,1): $z = 2(2) + 3(1) = 4 + 3 = 7$z=2(2)+3(1)=4+3=7

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The maximum value of $z$z is $9$9, occurring at the point $(0, 3)$(0,3).