A patient needs at least 40 units of vitamins and at least 30 units of iron a day from two foods F1 and F2. One unit of F1 contains 5 units of vitamins and 2 units of iron, and one unit of F2 contains 2 units of vitamins and 3 units of iron. One unit of F1 costs Rs. 8 and one unit of F2 costs Rs. 6. If x units of F1 and y units of F2 are used, the linear programming problem for the least cost is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Minimise \(Z = 8x + 6y\) subject to \(5x + 2y \ge 40\), \(2x + 3y \ge 30\), \(x \ge 0\), \(y \ge 0\)
The cost of x units of F1 and y units of F2 is the objective function, to be minimised:
\(Z = 8x + 6y\)
The vitamins taken must be at least 40:
\(5x + 2y \ge 40\)
The iron taken must be at least 30:
\(2x + 3y \ge 30\)
The quantities cannot be negative, so \(x \ge 0\), \(y \ge 0\).
Why the other options are incorrect
Option (1) Minimise \(Z = 8x + 6y\) subject to \(5x + 2y \le 40\), \(2x + 3y \le 30\), \(x \ge 0\), \(y \ge 0\) — Incorrect. “At least” needs \(\ge\); with \(\le\) the patient could take nothing at all.
Option (2) Minimise \(Z = 6x + 8y\) subject to \(5x + 2y \ge 40\), \(2x + 3y \ge 30\), \(x \ge 0\), \(y \ge 0\) — Incorrect. The prices are swapped: F1, which is x, costs Rs. 8.
Option (4) Minimise \(Z = 8x + 6y\) subject to \(5x + 2y \ge 40\), \(3x + 2y \ge 30\), \(x \ge 0\), \(y \ge 0\) — Incorrect. F1 has 2 units and F2 has 3 units of iron, so the iron constraint is \(2x + 3y \ge 30\).