Discussion
A manufacturing unit produces two products P1 and P2. For each piece of P1 and P2, the table below provides quantities of materials M1, M2, and M3 required, and also the profit earned. The maximum quantity available per day for M1, M2 and M3 is also provided. The maximum possible profit per day is ₹__________.
| M1 | M2 | M3 | Profit per piece (₹) | |
|---|---|---|---|---|
| P1 | 2 | 2 | 0 | 150 |
| P2 | 3 | 1 | 2 | 100 |
| Maximum quantity available per day | 70 | 50 | 40 |
Answer: B
The maximum profit is ₹4000 per day.
Let \(x\) and \(y\) be the daily output of P1 and P2. The linear programme is:
\[ \begin{aligned} \max\ Z &= 150x + 100y \\ 2x + 3y &\le 70 \quad (M1) \\ 2x + y &\le 50 \quad (M2) \\ 2y &\le 40 \quad (M3) \end{aligned} \]
The optimum lies at a corner of the feasible region. The M1 and M2 lines meet where \(2y = 20\), so \(y = 10\), \(x = 20\), which also satisfies M3. Evaluating the corners:
- (25, 0), limited by M2: \(Z = 3750\)
- (20, 10), M1 and M2: \(Z = 3000 + 1000 = 4000\)
- (5, 20), M1 and M3: \(Z = 750 + 2000 = 2750\)
- (0, 20): \(Z = 2000\)
The maximum is 4000 at 20 pieces of P1 and 10 of P2. 5000 and 6000 would need more material than is available (for example, 6000 needs 40 pieces of P1, using 80 units of M2 against 50), and 3000 is below an achievable corner.
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