Step-by-Step Solution

Topic: Linear Programming & Feasible Regions

(I) EQUATION OF LINE CD

The line passes through intercepts C(0, 6) and D(12, 0).

Using Intercept Form \( \frac{x}{a} + \frac{y}{b} = 1 \):

\( \frac{x}{12} + \frac{y}{6} = 1 \)

Multiply entire equation by 12:

Equation: \( x + 2y = 12 \)

(II) EQUATION OF LINE EF

The line passes through intercepts E(0, 4) and F(5, 0).

\( \frac{x}{5} + \frac{y}{4} = 1 \)

Multiply entire equation by 20:

Equation: \( 4x + 5y = 20 \)

(III)(A) DETERMINING CONSTRAINTS

  • 1. Line EF (4x + 5y = 20): Shaded region is away from origin (above).
    Constraint: \( 4x + 5y \ge 20 \)
  • 2. Line CD (x + 2y = 12): Shaded region is towards origin (below).
    Constraint: \( x + 2y \le 12 \)
  • 3. Line AB (passes 0,12 and 6,0): Equation is \( 2x + y = 12 \). Shaded region is below.
    Constraint: \( 2x + y \le 12 \)
  • 4. Non-negative Constraints: \( x \ge 0, y \ge 0 \)

(III)(B) MAXIMIZE Z = 600x + 400y

Step 1: Find Intersection Point X

Solve CD (\( x + 2y = 12 \)) and AB (\( 2x + y = 12 \)) simultaneously.

From AB: \( y = 12 – 2x \)

Substitute into CD: \( x + 2(12 – 2x) = 12 \)

\( x + 24 – 4x = 12 \)   ⇒   \( -3x = -12 \)   ⇒   \( x = 4 \)

\( y = 12 – 2(4) \)   ⇒   \( y = 4 \)

So, Point X is (4, 4).

Step 2: Evaluate Z at Corner Points

Corner Point Value of Z = 600x + 400y
C (0, 6) 2400
E (0, 4) 1600
F (5, 0) 3000
B (6, 0) 3600
X (4, 4) 600(4) + 400(4) = 4000 (Max)
Maximum Value of Z = 4000 at x=4, y=4

⚠️ Silly Mistake Audit: The “Inequality Direction”

A common error is confusing \( \le \) and \( \ge \). Remember the “Origin Test”: Pick (0,0). If the shaded region contains (0,0), the inequality must be true for 0 (e.g., \( 0 \le 12 \)). Since the region is above line EF, (0,0) is NOT included, so \( 4(0)+5(0) \ge 20 \) is false, confirming we use \( \ge \).

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