The corner points of the feasible region determined by the system of linear inequalities are (0, 0), (4, 0), (2, 4) and (0, 5). If the maximum value of Z = ax + by where \(a, b \gt 0\) occurs at both (2, 4) and (4, 0) then
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → a = 2b
The maximum occurs at both corners, so Z must have the same value at (2, 4) and (4, 0):
\(Z(2, 4) = 2a + 4b\)
\(Z(4, 0) = 4a\)
\(2a + 4b = 4a\)
\(4b = 2a\)
\(a = 2b\)
Check that this common value is the maximum: with a = 2b,
\(Z(4, 0) = Z(2, 4) = 8b\)
\(Z(0, 5) = 5b\)
\(Z(0, 0) = 0\)
Since b > 0, 8b is the largest value, so a = 2b.