Q.Solve the following L.P.P. by graphical method:
Minimize: $Z = 6x + 2y$ subject to $x + 2y \geq 3$, $x + 4y \geq 4$, $3x + y \geq 3$, $x \geq 0$, $y \geq 0$.

Evaluating $z = 3x + 2y$ at the corner points $(0,10)$, $(2,2)$ and $(4,0)$ gives $20$, $10$ and $12$; the least is $10$ at $(2,2)$.

By the corner-point (vertex) method, the optimum of a linear objective over a feasible polygon occurs at one of its corner points. Compute $z = 3x + 2y$ at each:

$z(0,10) = 3(0) + 2(10) = 20$

$z(2,2) = 3(2) + 2(2) = 6 + 4 = 10$ …

The optimum value of an LPP's objective function occurs at a corner point (vertex) of the feasible region, not at its centre. So the statement is False.

The objective function $Z=ax+by$ is linear, and the feasible region of an LPP is a convex polygon (bounded case). The corner-point theorem states that if an optimal solution exists, then the objective function attains its optimum value at one of the vertices (corner points) of the feasible region. The reason is that as $Z$ increases or decreases, its level lines sweep across the region and last touch it at a corner, never settli …

Evaluate $z = 4x + 5y$ at the four corner points; the values are $0,\ 12,\ 13,\ \frac{35}{3}$, so the maximum is $13$ at $(2,1)$. Hence option (B).

The corner-point (vertex) method states that for a linear objective function on a bounded feasible region, the optimum occurs at one of the corner points. Evaluating $z = 4x + 5y$: