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 Rs. __________.

	M1	M2	M3	Profit per piece (₹)
P1	2	2	0	150
P2	3	1	2	100
Maximum quantity available per day	70	50	40

The correct option is 4000.

Step-by-Step Explanation:
Let us formulate the given manufacturing problem as a Linear Programming Problem (LPP).
Let $x_1$ be the number of pieces of product P1 produced per day.
Let $x_2$ be the number of pieces of product P2 produced per day.

1. Formulation of the Objective Function:
The profit earned from each piece of P1 is ₹150, and from each piece of P2 is ₹100.
We want to maximize the total daily profit, which is given by:

$$\mathrm{Maximize}Z=150x_1+100x_2$$

2. Formulation of the Constraints:
Based on the daily availability limits of the materials M1, M2, and M3, we write the following constraints:
For Material M1 (maximum 70 units available per day):

$$2x_1+3x_2\le 70$$

For Material M2 (maximum 50 units available per day):

$$2x_1+x_2\le 50$$

For Material M3 (maximum 40 units available per day):

$$0x_1+2x_2\le 40\Rightarrow x_2\le 20$$

Since production cannot be negative, we have non-negativity constraints:

$$x_1\ge 0,x_2\ge 0$$

3. Determining the Corner Points of the Feasible Region:
We can find the boundary lines and determine the corner points of the feasible region by finding their intersections:

Point A: Intersection of the y-axis ($x_1=0$) and the M3 boundary line ($x_2=20$).
This gives the corner point: (0, 20).

Point B: Intersection of the M3 boundary line ($x_2=20$) and the M1 boundary line ($2x_1+3x_2=70$).
Substituting $x_2=20$ into the equation:
$2x_1+3\left(20\right)=70\Rightarrow 2x_1=10\Rightarrow x_1=5$.
Let us check if (5, 20) satisfies the M2 constraint: $2\left(5\right)+20=30\le 50$ (satisfies).
This gives the corner point: (5, 20).

Point C: Intersection of the M1 boundary line ($2x_1+3x_2=70$) and the M2 boundary line ($2x_1+x_2=50$).
Subtracting the second equation from the first:
$(2x_1+3x_2)-(2x_1+x_2)=70-50$
$2x_2=20\Rightarrow x_2=10$.
Substituting $x_2=10$ into the M2 equation:
$2x_1+10=50\Rightarrow 2x_1=40\Rightarrow x_1=20$.
This point satisfies $x_2=10\le 20$.
This gives the corner point: (20, 10).

Point D: Intersection of the M2 boundary line ($2x_1+x_2=50$) and the x-axis ($x_2=0$).
$2x_1=50\Rightarrow x_1=25$.
Let us check if (25, 0) satisfies the M1 constraint: $2\left(25\right)+3\left(0\right)=50\le 70$ (satisfies).
This gives the corner point: (25, 0).

Point E: The origin: (0, 0).

4. Calculating the Profit at each Corner Point:
We substitute the coordinates of each corner point $(x_1,x_2)$ into the profit function $Z=150x_1+100x_2$:

• At (0, 0):
$Z=150\left(0\right)+100\left(0\right)=0$

• At (0, 20):
$Z=150\left(0\right)+100\left(20\right)=2000$

• At (5, 20):
$Z=150\left(5\right)+100\left(20\right)=750+2000=2750$

• At (20, 10):
$Z=150\left(20\right)+100\left(10\right)=3000+1000=4000$

• At (25, 0):
$Z=150\left(25\right)+100\left(0\right)=3750$

Comparing all these values, the maximum possible profit per day is ₹4000 (which is achieved by producing 20 units of P1 and 10 units of P2).