The optimal value of the objective function of the LPP, Minimize Z = 3x - 2y subject to constraints x + y ≥ 10, 3x + 5y ≥ 15, x ≥ 0, y ≥ 0, is equal to: Option 1: 30 Option 2: -20 Option 3: -6 Option 4: 15

Solution: To find the feasible region, we look at the boundaries: Constraint 1: x + y ≥ 10. The line x + y = 10 passes through (10, 0) and (0, 10). Since it is ≥, the region is above this line. Constraint 2: 3x + 5y ≥ 15. The line 3x + 5y = 15 passes through (5, 0) and (0, 3). Non-negativity: x ≥ 0, y ≥ 0 restricts us to the first quadrant. Observation: Notice that the line x + y = 10 is further from the origin than 3x + 5y = 15 in the first quadrant. Therefore, x + y ≥ 10 is the "stricter" constraint, making 3x + 5y ≥ 15 redundant. We also see that there is no bounded region in this question. Since we are minimizing and the feasible region is unbounded in the positive y direction: As y increases (e.g., at point (0, 100)), Z becomes 3(0) - 2(100) = -200. As y ∞, Z -∞. The objective function is unbounded. In standard LPP theory, this means there is no finite optimal (minimum) value. Hence, NTA dropped this question! No marks were awarded.

Algebra

Linear Programming

Hard

core

The optimal value of the objective function of the LPP, Minimize $Z = 3x - 2y$ subject to constraints $x + y \ge 10$ , $3x + 5y \ge 15$ , $x \ge 0$ , $y \ge 0$ , is equal to:

-20

-6

✅ Drop

To find the feasible region, we look at the boundaries:

Constraint 1: $x + y \ge 10$ . The line $x + y = 10$ passes through $(10, 0)$ and $(0, 10)$ . Since it is $\ge$ , the region is above this line.

Constraint 2: $3x + 5y \ge 15$ . The line $3x + 5y = 15$ passes through $(5, 0)$ and $(0, 3)$ .

Solution figure for CUET Mathematics 2025 13 May Shift 2 question 29 (Algebra)

Non-negativity: $x \ge 0, y \ge 0$ restricts us to the first quadrant.

Observation: Notice that the line $x + y = 10$ is further from the origin than $3x + 5y = 15$ in the first quadrant. Therefore, $x + y \ge 10$ is the "stricter" constraint, making $3x + 5y \ge 15$ redundant.

We also see that there is no bounded region in this question.

Since we are minimizing and the feasible region is unbounded in the positive $y$ direction:

As $y$ increases (e.g., at point $(0, 100)$ ), $Z$ becomes $3(0) - 2(100) = -200$ .

As $y \to \infty$ , $Z \to -\infty$ .

The objective function is unbounded. In standard LPP theory, this means there is no finite optimal (minimum) value.

Hence, NTA dropped this question! No marks were awarded.

2026: 26 May Shift 2

If the feasible region (shaded portion)of a LPP is as shown in the graph, then max $z = 50x + 15y$ is:

2022: 25 May Shift 1

The feasible region of an LPP Max $Z = 3x + 2y$ subject to $x \geq 0, y \geq 0, x - 2y \leq 3$ is:

2026: 19 May Shift 2

Refer to the LPP given by $Z = 2.5x + y$ , subject to constraints

$3x + y \le 12; 3y + x \le 12; x, y \ge 0$ :

In the given figure, If $Z_A$ = value of Z at A, $Z_C$ = value of Z at C and Max.(Z) = maximum value of Z, then which of the following relations between $Z_A$ , $Z_C$ and Max.(Z) are correct?

A. $2Z_A = 5$ Max.(Z)

B. $5Z_A = 2Z_C$

C. $Z_A = 6 + Z_C$

D. $6+Z_A = Z_C$

Choose the correct answer from the options given below:

2026: 26 May Shift 2

Consider an LPP:

Maximize $Z = 5x + y$

Subject to $x + 3y \le 12$

$3x + y \le 12$

and $x,y \ge 0$

then which of the following statements are TRUE ?

A. The corner points of the bounded feasible region are (0,0), (4,0), (3,3) and (0,4).

B. Maximum value of $z$ occurs at (3,3).

C. Maximum value of $z$ occurs at (4,0).

D. The maximum value of $z$ is 20.

Choose the correct answer from the options given below:

2025: 26 May Shift 2

The corner points of a bounded feasible region determined by the following system of linear inequalities $x + 3y \leq 60, x + y \geq 10$ , $x \leq y$ , $x \geq 0$ , $y \geq 0$ are (0,10), (5,5), (15, 15) and (0, 20). Let $z = 2px + qy$ , $p, q > 0$ . If maximum of z occurs at both (15, 15) and (0, 20), then the relation between p and q is

2025: 21 May Shift 1

In a linear programming problem(LPP), the maximum value of the objective function $Z = 2x + 5y$ subjected to the constraints:

$2x + 3y ≤ 6$

$2x + y ≤ 4$

$x, y ≥ 0$

The optimal value of the objective function of the LPP, Minimize Z=3x−2yZ = 3x - 2yZ=3x−2y subject to constraints x+y≥10x + y \ge 10x+y≥10, 3x+5y≥153x + 5y \ge 153x+5y≥15, x≥0x \ge 0x≥0, y≥0y \ge 0y≥0, is equal to:

Solution

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CUET Mathematics Past Year Questions (Topic-Wise)

The optimal value of the objective function of the LPP, Minimize $Z = 3x - 2y$ subject to constraints $x + y \ge 10$ , $3x + 5y \ge 15$ , $x \ge 0$ , $y \ge 0$ , is equal to: