The corner points of the bounded feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let \(z = px + qy\) where \(p, q \gt 0\). Then the condition on \(p\) and \(q\) so that the maximum value of \(z\) occurs at (15, 15) and (0, 20) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(q = 3p\)
The values of z at the corner points are
\(z(0, 10) = 10q\)
\(z(5, 5) = 5p + 5q\)
\(z(15, 15) = 15p + 15q\)
\(z(0, 20) = 20q\)
The maximum occurs at both (15, 15) and (0, 20) only if the values there are equal:
\(15p + 15q = 20q\)
\(15p = 5q\)
\(q = 3p\)
With q = 3p the four values are
\(z(0, 10) = 30p,\quad z(5, 5) = 20p,\quad z(15, 15) = z(0, 20) = 60p\)
Since p > 0, 60p is the largest, so the maximum is reached at both points.
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Linear Programming
