Solved a Graphical method 1- linear programming (two | Chegg.com - Free Printable

Let's solve the given Linear Programming Problem (LPP) using the graphical method, as shown in your image.

---

📌 Problem Statement:

Minimize:
$$
Z = -3X_1 - 4X_2
$$

Subject to:
$$
\begin{align*}
(1)\quad & X_1 + X_2 \leq 20 \\
(2)\quad & -X_1 + 4X_2 \leq 20 \\
(3)\quad & X_1 \geq 10 \\
(4)\quad & X_2 \geq 5 \\
(5)\quad & X_1, X_2 \geq 0
\end{align*}
$$

---

🔍 Step-by-Step Solution Using Graphical Method

We are minimizing $ Z = -3X_1 - 4X_2 $. Since the coefficients of $ X_1 $ and $ X_2 $ are negative, we want to maximize the values of $ X_1 $ and $ X_2 $ within the feasible region to make $ Z $ as negative as possible (i.e., minimize it).

---

✔ Step 1: Plot the Constraints

We'll find the boundary lines and shade the feasible region.

#### Constraint 1: $ X_1 + X_2 \leq 20 $

- Line: $ X_1 + X_2 = 20 $
- Intercepts:
- $ X_1 = 0 \Rightarrow X_2 = 20 $
- $ X_2 = 0 \Rightarrow X_1 = 20 $
- Feasible side: Below the line

#### Constraint 2: $ -X_1 + 4X_2 \leq 20 $

- Line: $ -X_1 + 4X_2 = 20 $
- Rearranged: $ 4X_2 = X_1 + 20 \Rightarrow X_2 = \frac{1}{4}X_1 + 5 $
- Intercepts:
- $ X_1 = 0 \Rightarrow X_2 = 5 $
- $ X_2 = 0 \Rightarrow -X_1 = 20 \Rightarrow X_1 = -20 $ (not relevant since $ X_1 \geq 0 $)
- Feasible side: Below the line

#### Constraint 3: $ X_1 \geq 10 $

- Vertical line at $ X_1 = 10 $
- Feasible side: To the right

#### Constraint 4: $ X_2 \geq 5 $

- Horizontal line at $ X_2 = 5 $
- Feasible side: Above

#### Constraint 5: $ X_1, X_2 \geq 0 $

- First quadrant only

---

✔ Step 2: Identify the Feasible Region

The feasible region is bounded by all constraints. From the graph:

- The feasible region is a polygon with corner points: P, Q, R, S

From the table provided:

| Point | $ X_1, X_2 $ | $ Z = -3X_1 - 4X_2 $ |
|-------|----------------|------------------------|
| Q | (10, 7.5) | $ -3(10) - 4(7.5) = -30 - 30 = -60 $ |
| P | (10, 5) | $ -3(10) - 4(5) = -30 - 20 = -50 $ |
| R | (12, 8) | $ -3(12) - 4(8) = -36 - 32 = -68 $ |
| S | (15, 5) | $ -3(15) - 4(5) = -45 - 20 = -65 $ |

---

✔ Step 3: Evaluate Objective Function at Each Corner Point

We compute $ Z $ at each vertex:

- At Q (10, 7.5): $ Z = -60 $
- At P (10, 5): $ Z = -50 $
- At R (12, 8): $ Z = -68 $
- At S (15, 5): $ Z = -65 $

---

✔ Step 4: Find Minimum Value

We are minimizing $ Z $, so we look for the smallest (most negative) value.

- $ -68 < -65 < -60 < -50 $

So, the minimum value of $ Z $ is $ -68 $ at point R (12, 8).

---

✔ Final Answer:

$$
\boxed{
\text{Minimum } Z = -68 \text{ at } (X_1, X_2) = (12, 8)
}
$$

---

🔎 Verification of Point R (12, 8):

Check if it satisfies all constraints:

1. $ X_1 + X_2 = 12 + 8 = 20 \leq 20 $ → ✔
2. $ -X_1 + 4X_2 = -12 + 32 = 20 \leq 20 $ → ✔
3. $ X_1 = 12 \geq 10 $ → ✔
4. $ X_2 = 8 \geq 5 $ → ✔
5. Non-negativity → ✔

All satisfied. So, R is feasible.

---

📌 Conclusion:

✔ The optimal solution is at point R (12, 8), where the objective function reaches its minimum value of $ -68 $.

This matches the conclusion shown in the image.

---

✔ Answer:
$$
\boxed{\text{Min. } Z = -68 \text{ at } (X_1, X_2) = (12, 8)}
$$