The corner points of the bounded feasible region determined by a set of constraints (linear inequalities) are A(0, 5), B(3, 5), C(4, 0) and D(4, 1) and the objective function is \(z = px + 2qy\) where \(p, q \gt 0\). The condition on \(p\) and \(q\), such that, the maximum \(z\) occurs at B and D, is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(p-8q=0\)
Values of z at the corner points:
\(z(A) = 10q\)
\(z(B) = 3p + 10q\)
\(z(C) = 4p\)
\(z(D) = 4p + 2q\)
For the maximum to occur at both B and D, the values at B and D must be equal:
\(3p + 10q = 4p + 2q\)
\(p = 8q\), that is, \(p - 8q = 0\)
Check that this common value is the largest: with \(p = 8q\),
\(z(B) = z(D) = 34q,\quad z(C) = 32q,\quad z(A) = 10q\)
So the maximum, 34q, occurs at B and D.
Note: It seems that there is a mistake in the original NTA question, which gives the corner C as (5, 0). With that corner, \(p = 8q\) gives \(z(C) = 40q\), which is larger than \(z(B) = z(D) = 34q\), so the maximum could never occur at B and D. Also, D(4, 1) lies on the segment AC, so it could not be a corner. For the benefit of students and to ensure that the question is correct and has a single valid answer, the corner C has been changed to (4, 0), so that Option (1) gives the maximum at B and D.