Step-by-Step Solution
Topic: Graphical Constraints & Optimization
(I) TRIVIAL CONSTRAINTS
In most Linear Programming Problems involving physical quantities or number theory, the variables cannot be negative.
The trivial constraints are:
x ≥ 0, y ≥ 0
(II) NON-TRIVIAL CONSTRAINTS
Identify the Lines:
- • Line 1 (Points D & E): Intercepts (100,0) & (0,200) ⇒ 2x + y = 200
- • Line 2 (Points A & E): Intercepts (100,0) & (0,50) ⇒ x + 2y = 100 (Given)
- • Line 3 (Points O, B, C): Origin to (50,100) ⇒ y = 2x ⇒ 2x – y = 0
(a) For Region R1 (Quadrilateral ABCD):
R1 is below Line 1, above Line 2, and to the left of Line 3 (towards y-axis).
- 1. Below 2x + y = 200 ⇒ 2x + y ≤ 200
- 2. Left of y = 2x ⇒ y ≥ 2x (or 2x – y ≤ 0)
(b) For Region R2 (Triangle BCE):
R2 is below Line 1, above Line 2, and to the right of Line 3 (towards x-axis).
- 1. Below 2x + y = 200 ⇒ 2x + y ≤ 200
- 2. Right of y = 2x ⇒ y ≤ 2x (or 2x – y ≥ 0)
(III) MAXIMIZE z = 5x + 2y (On Region R1)
We evaluate Z at the corner points of R1: A, B, C, D.
| Corner Point | Value of Z = 5x + 2y |
| A (0, 50) | 5(0) + 2(50) = 100 |
| B (20, 40) | 5(20) + 2(40) = 100 + 80 = 180 |
| C (50, 100) | 5(50) + 2(100) = 250 + 200 = 450 (Max) |
| D (0, 200) | 5(0) + 2(200) = 400 |
Maximum Value = 450