For a linear programming problem, the feasible region is shown in the figure by shaded portion, then linear constraints are

The feasible region is in the first quadrant where both x and y are positive.

This gives: $x \geq 0, y \geq 0$

This eliminates Option 1 (which has $x \leq 0, y \leq 0$).

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The feasible region is bounded by two lines:

Line 1: $3x + 4y = 24$

Line 2: $x + 2y = 10$

Finding where the lines meet the axes:

For $3x + 4y = 24$:

When $x = 0$: $4y = 24$, so $y = 6$

When $y = 0$: $3x = 24$, so $x = 8$

For $x + 2y = 10$:

When $x = 0$: $2y = 10$, so $y = 5$

When $y = 0$: $x = 10$

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Using the origin $(0,0)$ as a test point to determine the inequalities:

For line $3x + 4y = 24$:

At origin: $3(0) + 4(0) = 0$

Since $0 < 24$ and the origin is inside the shaded region relative to this line:

$3x + 4y \leq 24$

For line $x + 2y = 10$:

At origin: $0 + 2(0) = 0$

Since $0 < 10$ but the origin is outside the shaded region relative to this line:

$x + 2y \geq 10$

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The complete set of constraints:

$3x + 4y \leq 24$

$x + 2y \geq 10$

$x \geq 0, y \geq 0$

This matches Option 4.