Maximizing Profits with Linear Programming: A Step-by-Step Guide Using the WorldLight Case Study
- vikash Singh
- Nov 7
- 3 min read
Learn how to use the graphical method to solve real-world business problems involving resource constraints and profit maximization.
Have you ever faced a situation where your resources are limited and you have to decide how much of each product to make to maximize profit?
That's exactly the problem faced by WorldLight Company — a manufacturer of light fixtures.
In this tutorial, we’ll walk you through how to use Linear Programming (LP) and the graphical method to solve this classic optimization problem — step by step.
🏭 Problem Statement
WorldLight Company manufactures two products:
Product 1 requires:
1 unit of metal frame
3 units of electrical components
Profit per unit: $1
Product 2 requires:
2 units of metal frame
2 units of electrical components
Profit per unit: $2 (only up to 60 units, no profit beyond that)
🔧 Resources available:
Metal frames: 200 units
Electrical components: 300 units
🎯 Objective: Maximize profit under resource constraints.
🔢 Step 1: Define Decision Variables
Let:
( x ) = number of Product 1 units to produce
( y ) = number of Product 2 units to produce
🧮 Step 2: Write the Objective Function
Maximize:
[Z = 1x + 2y]
(We’ll include the condition that ( y <= 60 ))
🔗 Step 3: Write the Constraints
From the problem:
Metal Frames:[1x + 2y <= 200]
Electrical Components:[3x + 2y <= 300]
Product 2 Limit:[y <= 60]
Non-negativity:[x >=0, y >= 0]
📉 Step 4: Graphical Solution
We graph the constraints to find the feasible region.
Convert constraints into equations:
( x + 2y = 200 ) → (Intercepts: ( (200, 0), (0, 100) ))
( 3x + 2y = 300 ) → (Intercepts: ( (100, 0), (0, 150) ))
( y = 60 ) (horizontal line)
Identify intersection points:
Intersect ( x + 2y = 200 ) and ( y = 60 ):
( x + 120 = 200 ) → ( x = 80 ), so point = ( (80, 60) )
Intersect ( 3x + 2y = 300 ) and ( y = 60 ):
( 3x + 120 = 300 ) → ( x = 60 ), so point = ( (60, 60) )
Intersect ( x + 2y = 200 ) and ( 3x + 2y = 300 ):
Subtract equations: ( (3x - x) = 100 ) → ( x = 50 )
Plug into first: ( 50 + 2y = 200 ) → ( y = 75 ) → point = ( (50, 75) )
✅ Feasible Corner Points:
( A = (0, 0) )
( B = (0, 60) )
( C = (60, 60) )
( D = (80, 60) )
( E = (50, 75) )
💰 Step 5: Calculate Profit at Each Feasible Point
Use ( Z = x + 2y ):
Point | ( x ) | ( y ) | ( Z = x + 2y ) |
A | 0 | 0 | 0 |
B | 0 | 60 | 120 |
C | 60 | 60 | 180 |
D | 80 | 60 | 200 (invalid as it violates the inequality 3x + 2y <= 300) |
E | 50 | 75 | 200 (invalid as y > 60) |
🏆 Step 6: Final Solution
Optimal Solution:
Produce 60 units of Product 1
Produce 60 units of Product 2
Maximum Profit = $180
The optimal point is at (60, 60).
This is it. Hope you found this useful.
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