Graphical Method for Solving L.P.P. (in Hindi) - Free Printable

The image shows a Linear Programming Problem (LPP) solved using the graphical method. Let's go through the problem step-by-step and verify the solution.

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🔷 Problem Statement:

Maximize:
$$
Z = 6x_1 + 11x_2
$$

Subject to:
$$
\begin{align*}
2x_1 + x_2 &\leq 104 \quad \text{(Constraint 1)} \\
x_1 + 2x_2 &\leq 76 \quad \text{(Constraint 2)} \\
x_1, x_2 &\geq 0 \quad \text{(Non-negativity)}
\end{align*}
$$

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🔷 Step-by-Step Solution:

#### ✔ Step 1: Convert Inequalities to Equations
We consider the constraints as equations to plot lines:

1. $ 2x_1 + x_2 = 104 $
2. $ x_1 + 2x_2 = 76 $

We find intercepts for each line.

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#### 📌 Line 1: $ 2x_1 + x_2 = 104 $

| $ x_1 $ | $ x_2 $ |
|----------|----------|
| 0 | 104 |
| 52 | 0 |

So, points: $ (0, 104) $, $ (52, 0) $

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#### 📌 Line 2: $ x_1 + 2x_2 = 76 $

| $ x_1 $ | $ x_2 $ |
|----------|----------|
| 0 | 38 |
| 76 | 0 |

So, points: $ (0, 38) $, $ (76, 0) $

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#### ✔ Step 2: Plot the Lines and Shade Feasible Region

We plot both lines on the graph and shade the region that satisfies all constraints (including non-negativity).

- The feasible region is bounded by:
- $ x_1 \geq 0 $
- $ x_2 \geq 0 $
- Below both lines.

From the graph:
- The feasible region is a polygon with corner points:
- $ A = (0, 0) $
- $ B = (0, 38) $ → from intersection of $ x_1 = 0 $ and $ x_1 + 2x_2 = 76 $
- $ C = (44, 16) $ → Intersection of two constraints
- $ D = (52, 0) $ → from $ x_2 = 0 $ and $ 2x_1 + x_2 = 104 $
- But wait — let’s check which points are actually vertices.

Wait! Let's verify the intersection point of the two lines.

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#### ✔ Step 3: Find Intersection Point of Two Constraints

Solve:
$$
\begin{cases}
2x_1 + x_2 = 104 \quad \text{(1)}\\
x_1 + 2x_2 = 76 \quad \text{(2)}
\end{cases}
$$

Use substitution or elimination.

Multiply equation (2) by 2:
$$
2x_1 + 4x_2 = 152 \quad \text{(2a)}
$$

Subtract equation (1) from (2a):
$$
(2x_1 + 4x_2) - (2x_1 + x_2) = 152 - 104 \\
3x_2 = 48 \Rightarrow x_2 = 16
$$

Plug back into (2):
$$
x_1 + 2(16) = 76 \Rightarrow x_1 = 76 - 32 = 44
$$

✔ So, intersection point: $ (44, 16) $

Now list all corner points of the feasible region:

1. $ (0, 0) $
2. $ (0, 38) $ → from $ x_1 = 0 $ and $ x_1 + 2x_2 = 76 $
3. $ (44, 16) $ → intersection of two constraints
4. $ (52, 0) $ → from $ x_2 = 0 $ and $ 2x_1 + x_2 = 104 $

But wait — we must check if all these points satisfy both constraints.

Let’s verify:

- $ (0, 0) $: OK
- $ (0, 38) $: Check $ 2(0) + 38 = 38 \leq 104 $? Yes. $ 0 + 2(38) = 76 \leq 76 $? Yes → Valid
- $ (52, 0) $: $ 2(52) + 0 = 104 \leq 104 $? Yes. $ 52 + 0 = 52 \leq 76 $? Yes → Valid
- $ (44, 16) $: Already derived from both equations → Valid

So, the feasible region has vertices:
- $ (0, 0) $
- $ (0, 38) $
- $ (44, 16) $
- $ (52, 0) $

But wait — does the line from $ (0, 38) $ to $ (44, 16) $ lie along $ x_1 + 2x_2 = 76 $? Yes.

And from $ (44, 16) $ to $ (52, 0) $ lies along $ 2x_1 + x_2 = 104 $? Let's check:

At $ (52, 0) $: $ 2(52) + 0 = 104 $ → yes.

So the feasible region is a quadrilateral with vertices:
- $ A = (0, 0) $
- $ B = (0, 38) $
- $ C = (44, 16) $
- $ D = (52, 0) $

But now, is (0, 38) connected directly to (44,16)? Yes — because that's where $ x_1 + 2x_2 = 76 $ cuts off.

Similarly, $ (44,16) $ to $ (52,0) $ is along $ 2x_1 + x_2 = 104 $

So yes — four corner points.

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#### ✔ Step 4: Evaluate Objective Function at Each Corner

We compute $ Z = 6x_1 + 11x_2 $ at each vertex.

| Point | $ x_1 $ | $ x_2 $ | $ Z = 6x_1 + 11x_2 $ |
|-----------|---------|---------|------------------------|
| (0, 0) | 0 | 0 | $ 0 $ |
| (0, 38) | 0 | 38 | $ 6(0) + 11(38) = 418 $ |
| (44, 16) | 44 | 16 | $ 6(44) + 11(16) = 264 + 176 = 440 $ |
| (52, 0) | 52 | 0 | $ 6(52) = 312 $ |

✔ Maximum $ Z = 440 $ at $ (44, 16) $

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✔ Final Answer:

$$
\boxed{
\text{Maximum } Z = 440 \text{ at } x_1 = 44, \; x_2 = 16
}
$$

This matches what is written in the image.

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🔍 Verification of Graph:

- The shaded region is correct: bounded by:
- $ x_1 = 0 $
- $ x_2 = 0 $
- $ 2x_1 + x_2 = 104 $
- $ x_1 + 2x_2 = 76 $
- Points labeled:
- $ (0, 104) $: on first line, but not feasible since $ x_1 + 2x_2 = 104 > 76 $
- $ (0, 38) $: valid
- $ (44, 16) $: intersection
- $ (52, 0) $: valid
- $ (76, 0) $: outside constraint $ 2x_1 + x_2 \leq 104 $? $ 2(76) = 152 > 104 $ → not feasible

So $ (76, 0) $ should not be in feasible region — but it's marked on the graph.

⚠️ Note: The graph shows $ (76, 0) $ on the axis, but this point violates the first constraint:
$ 2(76) + 0 = 152 > 104 $ → Not feasible

So the correct feasible region ends at (52, 0), not (76, 0)

✔ The shaded region in the image seems to go only up to $ (52, 0) $, so it's correct.

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✔ Conclusion:

Your solution is correct.

- The optimal solution is:
$$
\boxed{x_1 = 44,\quad x_2 = 16,\quad \text{Max } Z = 440}
$$

This is achieved at the intersection of the two constraints.

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💡 Summary of Steps:

1. Convert inequalities to equations.
2. Find intercepts and plot lines.
3. Identify feasible region (bounded by constraints).
4. Find corner points.
5. Evaluate objective function at each corner.
6. Choose maximum value.

✔ This is the standard graphical method for solving LPPs with two variables.

Let me know if you want the solution in LaTeX or need help plotting it!