Step-by-Step Solution

Topic: Linear Programming Optimization

(I) UNITS FROM FACTORY P TO DEPOT C

Factory P has a total capacity of 9 units.

• Sent to A = \( x \)
• Sent to B = \( y \)

Remaining units must go to C:

(II) UNITS FROM FACTORY Q

Factory Q fills the remaining demand for each depot.

• To A: Depot A needs 4. Received \( x \) from P.
  Q → A = \( 4 – x \)
• To B: Depot B needs 4. Received \( y \) from P.
  Q → B = \( 4 – y \)
• To C: Depot C needs 6. Received \( 9 – x – y \) from P.
  Q → C = \( 6 – (9 – x – y) \)
  Q → C = \( x + y – 3 \)

(III)(A) TOTAL COST FUNCTION Z

Multiply quantity by cost per unit for all 6 routes:

\( Z = 160x + 100y + 150(9 – x – y) + 100(4 – x) + 120(4 – y) + 100(x + y – 3) \)

Simplify:

\( Z = 160x + 100y + 1350 – 150x – 150y + 400 – 100x + 480 – 120y + 100x + 100y – 300 \)

Combine like terms:

(III)(B) CONSTRAINT INEQUALITIES

All quantities transported must be non-negative (\( \ge 0 \)).

1. \( x \ge 0 \) and \( y \ge 0 \)
2. \( 9 – x – y \ge 0 \)   ⇒   \( x + y \le 9 \)
3. \( 4 – x \ge 0 \)   ⇒   \( x \le 4 \)
4. \( 4 – y \ge 0 \)   ⇒   \( y \le 4 \)
5. \( x + y – 3 \ge 0 \)   ⇒   \( x + y \ge 3 \)