Linear Programming — Class 12 Maths Solution

From the figure, it is clear that feasible region is bounded with

coordinates of corner points as (0,3),(3,2) and (0,5).

Here, $Z = 11x + 7y$.

$x + 3y = 9$ and $x + y = 5$

$\Rightarrow$ $2y = 4$

$\therefore y = 2$ and $x = 3$

So, intersection points of $x + y = 5$ and $x + 3y = 9$ is (3,2).

Hence, the minimum value of Z is 21 at (0,3).