Linear inequalities corresponding to the shaded feasible region OABCO in the given figure are Option 1: 4x + 3y ≤ 12, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0 Option 2: 4x + 3y ≥ 12, 3x + 4y ≥ 12, x ≥ 0, y ≥ 0 Option 3: 4x + 3y ≤ 12, 3x + 4y ≥ 12, x ≥ 0, y ≥ 0 Option 4: 4x + 3y ≥ 12, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0

The answer is '4x + 3y ≤ 12, 3x + 4y ≤ 12, x ≥ 0, y ≥ 0'. View full solution:

Linear inequalities corresponding to the shaded feasib<= region OABCO in the given figure are

4x + 3y \>=q 12, 3x + 4y \>=q 12, x \>=q 0, y \>=q 0

4x + 3y \ =q 12, x \>=q 0, y \>=q 0

4x + 3y \>=q 12, 3x + 4y \ =q 0, y \>=q 0

Algebra > Linear Programming

Medium

common

Linear inequalities corresponding to the shaded feasible region OABCO in the given figure are <img src="https://balti.afterboards.in/XDcoYd2BxodBB7e" width="300px"/>

$4x + 3y \leq 12, 3x + 4y \leq 12, x \geq 0, y \geq 0$

$4x + 3y \geq 12, 3x + 4y \geq 12, x \geq 0, y \geq 0$

$4x + 3y \leq 12, 3x + 4y \geq 12, x \geq 0, y \geq 0$

$4x + 3y \geq 12, 3x + 4y \leq 12, x \geq 0, y \geq 0$

✅ Correct Option: 1

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Linear inequalities corresponding to the shaded feasible region OABCO in the given figure are <img src="https://balti.afterboards.in/XDcoYd2BxodBB7e" width="300px"/>

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