For the L.P.P. Maximize z = 10x + 6y subjected to

3x + y ≤ 12,

2x + 5y ≤ 34,

x, y ≥ 0.

Then the feasible region represented by system of inequalities is

The given constraints are:

$3x + y \leq 12$

$2x + 5y \leq 34$

$x, y \geq 0$

The conditions $x \geq 0$ and $y \geq 0$ restrict the region to the first quadrant.

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To find the boundary lines, convert inequalities to equations:

For $3x + y = 12$:

When $x = 0$: $y = 12$ giving point $(0, 12)$

When $y = 0$: $x = 4$ giving point $(4, 0)$

For $2x + 5y = 34$:

When $x = 0$: $y = 6.8$ giving point $(0, 6.8)$

When $y = 0$: $x = 17$ giving point $(17, 0)$

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To find the intersection point, solve:

$3x + y = 12$ ... (i)

$2x + 5y = 34$ ... (ii)

From equation (i):

$y = 12 - 3x$

Substituting in equation (ii):

$2x + 5(12 - 3x) = 34$

$2x + 60 - 15x = 34$

$-13x = -26$

$x = 2$

Substituting $x = 2$ in equation (i):

$y = 12 - 3(2)$

$y = 6$

The intersection point is $(2, 6)$.

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The feasible region is the area satisfying all constraints:

• Below line $3x + y = 12$
• Below line $2x + 5y = 34$
• In the first quadrant

Corner points of the feasible region:

$(0, 0)$, $(4, 0)$, $(2, 6)$, $(0, 6.8)$

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The feasible region is enclosed by two constraint lines that intersect, along with the x-axis and y-axis. This forms a closed polygon that does not extend to infinity.

Therefore, the feasible region is bounded in the first quadrant.