✅ Why do we need Linear Programming?

To optimize use of limited resources To help in strategic decision-making Common in manufacturing, transportation, finance, and logistics

📦 Example Problem: Chocolate Manufacturing Company

🟤 Produces Two Chocolates: A and B

Each unit of A and B requires:

• Milk and Cocoa (quantities shown below)

Chocolate	Milk (units)	Cocoa (units)	Profit per unit
A	1	3	₹6
B	1	2	₹5

Total resources available:

• Milk: 5 units

• Cocoa: 12 units

✅ Step 1: Define Variables

Let:

• x = number of units of chocolate A

• y = number of units of chocolate B

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✅ Step 2: Objective Function

Maximize Profit:

Z=6x+5y

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✅ Step 3: Constraints

1. Milk constraint: x+y ≤ 5

2. Cocoa constraint: 3x+2y ≤ 12

3. Non-negativity: x≥0, y≥0

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✅ Step 4: Find Corner Points

Let’s solve the system manually by checking intercepts and intersections:

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🔹 Constraint 1: x+y=5

• If x=0 , then y=5

• If y=0, then x=5

Line passes through (0, 5) and (5, 0)

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🔹 Constraint 2: 3x+2y=12

• If x=0x = 0x=0, then 2y=12⇒y=62y = 12 \Rightarrow y = 62y=12⇒y=6

• If y=0y = 0y=0, then 3x=12⇒x=43x = 12 \Rightarrow x = 43x=12⇒x=4

Line passes through (0, 6) and (4, 0)

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🔹 Find Point of Intersection

Solve:

1. x+y=5

2. 3x+2y=12

Substitute y=5−x into 2:

3x+2(5−x)=123x+10−2x=12x=2⇒y=33x + 2(5 - x) = 12 \\ 3x + 10 - 2x = 12 \\ x = 2 \Rightarrow y = 33x+2(5−x)=123x+10−2x=12x=2⇒y=3

🟢 Intersection point: (2,3)

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✅ Step 5: Evaluate Profit at Feasible Points

Check all corner points of the feasible region:

Point	Check x+y≤5x + y \leq 5x+y≤5	3x+2y≤123x + 2y \leq 123x+2y≤12	Z=6x+5yZ = 6x + 5yZ=6x+5y
(0, 0)	✅	✅	0
(0, 5)	✅	3(0)+2(5)=10≤123(0) + 2(5) = 10 \leq 123(0)+2(5)=10≤12 ✅	25
(5, 0)	✅	3(5)=15>123(5) = 15 > 123(5)=15>12 ❌	-
(4, 0)	✅	✅	24
(2, 3)	✅	✅	6×2+5×3=12+15=276×2 + 5×3 = 12 + 15 = 276×2+5×3=12+15=27 ✅

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✅ Final Answer:

• Optimal solution: x=2, y=3x = 2,\ y = 3x=2, y=3

• Produce:

  • 2 units of Chocolate A

  • 3 units of Chocolate B

• Maximum Profit: ₹27

// Some code
import matplotlib.pyplot as plt
import numpy as np

# Define new constraints
x = np.linspace(0, 10, 400)
y1 = 5 - x           # From x + y <= 5 (milk constraint)
y2 = (12 - 3*x) / 2  # From 3x + 2y <= 12 (cocoa constraint)

# Setup plot
plt.figure(figsize=(8, 6))

# Plot constraint lines
plt.plot(x, y1, label=r'$x + y \leq 5$', color='blue')
plt.plot(x, y2, label=r'$3x + 2y \leq 12$', color='green')

# Fill feasible region
y_feasible = np.minimum(y1, y2)
plt.fill_between(x, 0, y_feasible, where=(y_feasible >= 0), color='skyblue', alpha=0.5, label='Feasible Region')

# Plot feasible corner points
points = {
    "A (0,0)": (0, 0),
    "B (0,5)": (0, 5),
    "C (4,0)": (4, 0),
    "D (2,3)": (2, 3),  # Intersection point
}
for label, (px, py) in points.items():
    plt.plot(px, py, 'ro')
    plt.text(px + 0.2, py + 0.2, label, fontsize=10)

# Highlight optimal point (2, 3)
plt.plot(2, 3, 'ko', markersize=8, label='Optimal Solution (2, 3)')

# Chart decorations
plt.xlim(0, 8)
plt.ylim(0, 8)
plt.xlabel('Units of Chocolate A (x)')
plt.ylabel('Units of Chocolate B (y)')
plt.title('Feasible Region - Chocolate Production (Updated Constraints)')
plt.grid(True)
plt.legend()
plt.tight_layout()
plt.show()