Which of the region shown in the given figures represents the feasible region bounded by the following constraints?

$4x + y \geq 80$, $2x + y \geq 60$, $x + y \leq 80$, $x \geq 0$, $y \geq 0$

The feasible region satisfies all the given constraints simultaneously: $4x + y \geq 80$, $2x + y \geq 60$, $x + y \leq 80$, $x \geq 0$, $y \geq 0$.

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The boundary lines are:

Line 1: $4x + y = 80$

Line 2: $2x + y = 60$

Line 3: $x + y = 80$

Lines 4 & 5: $x = 0$ (y-axis) and $y = 0$ (x-axis)

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For Line 1: $4x + y = 80$

When $x = 0$: $y = 80$ giving point $(0, 80)$

When $y = 0$: $4x = 80$

$x = 20$ giving point $(20, 0)$

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For Line 2: $2x + y = 60$

When $x = 0$: $y = 60$ giving point $(0, 60)$

When $y = 0$: $2x = 60$

$x = 30$ giving point $(30, 0)$

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For Line 3: $x + y = 80$

When $x = 0$: $y = 80$ giving point $(0, 80)$

When $y = 0$: $x = 80$ giving point $(80, 0)$

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Testing the origin $(0, 0)$ in each inequality:

For $4x + y \geq 80$: $0 \geq 80$ is false, so the region is away from origin

For $2x + y \geq 60$: $0 \geq 60$ is false, so the region is away from origin

For $x + y \leq 80$: $0 \leq 80$ is true, so the region is towards origin

For $x \geq 0$ and $y \geq 0$: the region is in the first quadrant

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The feasible region must be:

• Above both $4x + y = 80$ and $2x + y = 60$
• Below $x + y = 80$
• In the first quadrant

Region C satisfies all these conditions. It lies between the two lower lines, stays below the upper line, and remains in the first quadrant.

Region A is below all lines and violates both $\geq$ constraints.

Region B is too low and violates $4x + y \geq 80$.

Region D is above $x + y = 80$ and violates $x + y \leq 80$.

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Therefore, the answer is Option 3 (Region C).