The solution set of the linear constraints x - 2y ≥ 0, 2x - y ≤ -4, x ≥ 0 and y ≥ 0 is Option 1: Option 2: \(-1,0), (0,2)\ Option 3: \(-1,0), (0,2), (0, 0)\ Option 4: unbounded region in first quadrant

Solution: The four constraints are: 1. x - 2y ≥ 0 2. 2x - y ≤ -4 3. x ≥ 0 4. y ≥ 0 The solution set contains points that satisfy all four constraints simultaneously. From constraint 1: x - 2y ≥ 0 x ≥ 2y ... (A) From constraint 2: 2x - y ≤ -4 -y ≤ -4 - 2x y ≥ 2x + 4 ... (B) Constraints 3 and 4 indicate the first quadrant. From (B): y ≥ 2x + 4 Substituting into (A): x ≥ 2y x ≥ 2(2x + 4) x ≥ 4x + 8 -3x ≥ 8 x ≤ -8/3 The constraints require: x ≤ -8/3 (approximately -2.67) AND x ≥ 0 (from constraint 3) These cannot be satisfied simultaneously. No point can have x ≤ -8/3 and x ≥ 0 at the same time. Since no point satisfies all four constraints, the solution set is empty. Therefore, the solution set is .

Algebra

Linear Programming

Medium

common

The solution set of the linear constraints $x - 2y \geq 0, 2x - y \leq -4, x \geq 0$ and $y \geq 0$ is

$\phi$

$\{(-1,0), (0,2)\}$

$\{(-1,0), (0,2), (0, 0)\}$

unbounded region in first quadrant

✅ Correct Option: 1

The four constraints are:

$x - 2y \geq 0$
$2x - y \leq -4$
$x \geq 0$
$y \geq 0$

The solution set contains points that satisfy all four constraints simultaneously.

From constraint 1: $x - 2y \geq 0$

$x \geq 2y$ ... (A)

From constraint 2: $2x - y \leq -4$

$-y \leq -4 - 2x$

$y \geq 2x + 4$ ... (B)

Constraints 3 and 4 indicate the first quadrant.

From (B): $y \geq 2x + 4$

Substituting into (A):

$x \geq 2y$

$x \geq 2(2x + 4)$

$x \geq 4x + 8$

$-3x \geq 8$

$x \leq -\dfrac{8}{3}$

The constraints require:

$x \leq -\dfrac{8}{3}$ (approximately $-2.67$ )

AND

$x \geq 0$ (from constraint 3)

These cannot be satisfied simultaneously. No point can have $x \leq -\dfrac{8}{3}$ and $x \geq 0$ at the same time.

Since no point satisfies all four constraints, the solution set is empty.

Therefore, the solution set is $\phi$ .

2026: 19 May Shift 1

The corner points O, A, B, C, D and E of the bounded feasible region determined by the system of linear constraints are shown in the given figure.

If $z = 3x - 4y$ be the objective function, then the minimum of z, is:

2025: 15 May Shift 1

Let the corner points of the bounded feasible region of the linear programming problem (LPP) $Z = ax+by$ be: (0, 0), (2, 0), (20/19, 45/19) and (0, 3). If the optimal value of Z occurs at both points (2, 0) and (20/19, 45/19), then the relation between a and b is:

2025: 14 May Shift 1

Which of the following is NOT a basic requirement of the linear programming problem (LPP)?

2023: 30 May Shift 3

The feasible region for a LPP is shown in the given figure. The maximum value of $z = 2x + 5y$ is :

2026: 18 May Shift 1

The corner points of the bounded feasible region for an L.P.P are (0, 10), (5, 5), (15,15), and (0, 20). If the objective function is $Z=px+qy, p,q>0$ , then the condition on $p$ and $q$ so that the maximum of $Z$ occurs at (15, 15) and (0, 20) 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 solution set of the linear constraints x−2y≥0,2x−y≤−4,x≥0x - 2y \geq 0, 2x - y \leq -4, x \geq 0x−2y≥0,2x−y≤−4,x≥0 and y≥0y \geq 0y≥0 is

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CUET Mathematics Past Year Questions (Topic-Wise)

The solution set of the linear constraints $x - 2y \geq 0, 2x - y \leq -4, x \geq 0$ and $y \geq 0$ is