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