A LPP max\ z=ax+by having bounded feasible region with vertices on (2…

The correct answer is Option (3) → (A) and (D) only

Going round the vertices in order, (5, 6) and (1, 7) are adjacent, so the segment joining them is an edge of the region. The values of z there are

    \( \displaystyle z(1,7)=a+7b \)
    \( \displaystyle z(5,6)=5a+6b \)

With b = 4a, \( z=a(x+4y) \), and the corner values are

    \( \displaystyle z(2,1)=6a \)
    \( \displaystyle z(3,2)=11a \)
    \( \displaystyle z(5,6)=29a \)
    \( \displaystyle z(1,7)=29a \)

Taking a > 0, as usual for a maximisation problem, the two largest values are equal.

With b = −4a,

    \( \displaystyle z(1,7)=a-28a=-27a \)
    \( \displaystyle z(5,6)=5a-24a=-19a \)

So with b = 4a the maximum 29a occurs at both (5, 6) and (1, 7); with b = −4a the values at these two points (−27a and −19a) are not equal; and the region is bounded.

Checking the statements

(A) \( \max\ z \) occurs at (1, 7) and (5, 6) if \( b=4a \) — Correct. z = 29a at both points, larger than 6a and 11a at the other corners.

(B) \( \max\ z \) occurs at (1, 7) and (5, 6) if \( b=-4a \) — Incorrect. z(1, 7) = −27a and z(5, 6) = −19a are not equal, so the maximum cannot occur at both points.

(C) \( \max\ z \) does not exist — Incorrect. The feasible region is bounded, so z always has a maximum, at a corner point.

(D) if \( \max\ z \) occurs at (1, 7) and (5, 6) , then \( \max\ z \) occurs for infinite number of value of \( x \) and \( y \) — Correct. When the maximum occurs at two adjacent corners, it occurs at every point of the edge joining them, and that segment has infinitely many points.