A housing society installs x solar units and y wind units. Each solar unit supplies 6 units of energy during the day and 1 unit at night, and each wind unit supplies 2 units during the day and 3 units at night. The society needs at least 24 units during the day and at least 12 units at night. A solar unit costs Rs. 5000 and a wind unit costs Rs. 4000. The least cost of meeting the needs is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Rs. 27000
The linear programming problem is
Minimise \(Z = 5000x + 4000y\)
subject to
\(6x + 2y \ge 24\), that is, \(3x + y \ge 12\)
\(x + 3y \ge 12\)
\(x \ge 0, \; y \ge 0\)
The lines \(3x + y = 12\) and \(x + 3y = 12\) meet where
\(x = 3, \; y = 3\)
The region is unbounded, with corner points (0, 12), (3, 3) and (12, 0). Values of Z:
\(Z(0, 12) = 48000\)
\(Z(3, 3) = 15000 + 12000 = 27000\)
\(Z(12, 0) = 60000\)
Since the region is unbounded, 27000 is the minimum only if no feasible point gives less. Every feasible point satisfies
\(8(5x + 4y) = 11(3x + y) + 7(x + 3y) \ge 132 + 84 = 216\)
\(5x + 4y \ge 27\)
So the least cost is Rs. 27000, with 3 solar and 3 wind units.
Why the other options are incorrect
Option (2) Rs. 24000 — Incorrect. Every feasible choice costs at least Rs. 27000.
Option (3) Rs. 30000 — Incorrect. The needs can be met for Rs. 27000, so this is not the least.
Option (4) Rs. 48000 — Incorrect. This is the cost with 12 wind units only, at (0, 12).