Consider the feasible region determined by the constraints \(x + y \ge 5\), \(x + 3y \ge 9\), \(x \ge 0\), \(y \ge 0\).
Match each objective function in List-I with its minimum value on this region in List-II.
| List-I | List-II | ||
| (A) | \(Z = 3x + y\) | (I) | 9 |
| (B) | \(Z = x + 2y\) | (II) | 12 |
| (C) | \(Z = x + 5y\) | (III) | 5 |
| (D) | \(Z = 2x + 3y\) | (IV) | 7 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The region is unbounded. The line x + y = 5 meets the y-axis at (0, 5), and x + 3y = 9 meets the x-axis at (9, 0). The two lines meet where
\(2y = 9 - 5 = 4\)
\(y = 2,\ x = 3\)
The corner points are (0, 5), (3, 2) and (9, 0). Each objective below has positive coefficients, and in each case the smallest corner value is also a lower bound for Z on the whole region, so it is the minimum.
(A) \(Z = 3x + y\)
This matches (III) 5.
\(Z(0, 5) = 5,\ Z(3, 2) = 11,\ Z(9, 0) = 27\)
\(3x + y \ge x + y \ge 5\)
(B) \(Z = x + 2y\)
This matches (IV) 7.
\(Z(0, 5) = 10,\ Z(3, 2) = 7,\ Z(9, 0) = 9\)
\(\displaystyle x + 2y = \frac{1}{2}(x + y) + \frac{1}{2}(x + 3y) \ge \frac{5}{2} + \frac{9}{2} = 7\)
(C) \(Z = x + 5y\)
This matches (I) 9.
\(Z(0, 5) = 25,\ Z(3, 2) = 13,\ Z(9, 0) = 9\)
\(x + 5y \ge x + 3y \ge 9\)
(D) \(Z = 2x + 3y\)
This matches (II) 12.
\(Z(0, 5) = 15,\ Z(3, 2) = 12,\ Z(9, 0) = 18\)
\(\displaystyle 2x + 3y = \frac{3}{2}(x + y) + \frac{1}{2}(x + 3y) \ge \frac{15}{2} + \frac{9}{2} = 12\)