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 ₹ ________________.
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For linear programming problems with two variables, check corner points formed by constraint intersections. The maximum profit always lies at a feasible corner.
Let $x$ be the number of P1 units produced and $y$ be the number of P2 units produced.
Material constraints are:
\[
\text{M1: } 2x + 3y \le 70,
\qquad
\text{M2: } 2x + y \le 50,
\qquad
\text{M3: } 2y \le 40 \Rightarrow y \le 20.
\]
Profit function:
\[
Z = 150x + 100y.
\]
Use the binding constraint $y = 20$ (since M3 is tight).
Substitute in M1 and M2:
\[
2x + 3(20) \le 70 \Rightarrow 2x \le 10 \Rightarrow x \le 5,
\]
\[
2x + 20 \le 50 \Rightarrow x \le 15.
\]
So the limiting value is $x = 5$.
Profit:
\[
Z = 150(5) + 100(20)
= 750 + 2000
= 2750.
\]
But check if $y = 0$ gives higher profit (profit per M1 is better for P1):
\[
2x \le 70 \Rightarrow x = 35 \quad\text{but M2: } 2x \le 50 \Rightarrow x \le 25.
\]
So $x = 25, y = 0$.
Profit:
\[
Z = 150(25) = 3750.
\]
Now check the intersection of M1 and M2 constraints:
Solve
\[
2x + 3y = 70,
\qquad
2x + y = 50.
\]
Subtract:
\[
2y = 20 \Rightarrow y = 10,
\]
\[
2x + 10 = 50 \Rightarrow x = 20.
\]
Profit:
\[
Z = 150(20) + 100(10) = 3000 + 1000 = 4000.
\]
This is the maximum feasible profit.
Final Answer: 4000
