The solution set of the linear constraints $x-2y≥ 0,2x-y≤-4,x ≥ 0$ and $y ≥ 0$ is

$\phi$

The correct answer is Option (1) → $\phi$

Given constraints:

1) $x - 2y \geq 0 \Rightarrow x \geq 2y$

2) $2x - y \leq -4$

3) $x \geq 0$, $y \geq 0$

Now analyze the feasible region step-by-step:

From (1): $x \geq 2y$ (Region lies to the right of the line $x = 2y$)

From (2): $2x - y \leq -4 \Rightarrow y \geq 2x + 4$ (Region lies above the line $y = 2x + 4$)

Also, $x \geq 0$ and $y \geq 0$ means we are in the first quadrant.

We combine these to get:

$x \geq 2y$ and $y \geq 2x + 4$

Substitute $x = 2y$ into $y \geq 2x + 4$:

$y \geq 2(2y) + 4 = 4y + 4$ ⟹ $y - 4y \geq 4$ ⟹ $-3y \geq 4$ ⟹ $y \leq -\frac{4}{3}$

But this contradicts $y \geq 0$.

Conclusion: No feasible region exists that satisfies all constraints.

Answer: The solution set is the empty set.