The corner points of feasible region determined by the following system of linear inequalities:
\(2x+y\le 10\), \(x+3y\le 15\), \(x\ge 0\), \(y\ge 0\) are (0, 0), (5, 0), (3, 4) and (0, 5).
Let \(z=px+qy\), where \(p, q\gt 0\), then the relation between p and q, so that maximum of z occurs at both points (3, 4) and (0, 5) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3p=q\)
If the maximum occurs at two corners, z must take the same value at both:
\(z(3, 4) = 3p+4q\)
\(z(0, 5) = 5q\)
\(3p+4q = 5q\)
\(3p = q\)
Check: with q = 3p, z(3, 4) = z(0, 5) = 15p, while z(5, 0) = 5p and z(0, 0) = 0 are smaller. So 15p is indeed the maximum.
NCERT Exemplar:
Linear Programming
