GATE 2022 ME (ME2) – Question 42
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 ₹__________.
| Product | M1 | M2 | M3 | Profit per piece (₹) |
|---|---|---|---|---|
| P1 | 2 | 2 | 0 | 150 |
| P2 | 3 | 1 | 2 | 100 |
| Maximum available per day | 70 | 50 | 40 |
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Correct answer: (B) 4000
Explanation
Let $x$ be the number of P1 and $y$ the number of P2 made per day.
**Maximise** $Z=150x+100y$
**subject to**
- M1: $2x+3y\leq70$
- M2: $2x+y\leq50$
- M3: $2y\leq40\;\Rightarrow\;y\leq20$
- $x,y\geq0$.
**Corner points of the feasible region.**
- $(0,0)$: $Z=0$.
- $(25,0)$ (M2 with $y=0$): $Z=3750$.
- $(0,20)$: $Z=2000$.
- M2 ∩ M3: $y=20$, $2x+20=50\Rightarrow x=15$, i.e. $(15,20)$. Check M1: $30+60=90>70$, so it is infeasible.
- M1 ∩ M2: subtract: $(2x+3y)-(2x+y)=70-50\Rightarrow2y=20$, $y=10$, $x=20$. At $(20,10)$: M3 gives $y=10\leq20$ ✓. $Z=150(20)+100(10)=4000$.
- M1 ∩ M3: $y=20$, $2x+60=70$, $x=5$, i.e. $(5,20)$. $Z=750+2000=2750$.
The largest value is at $(20,10)$:
$$Z_{max}=\mathbf{₹4000}.$$