Question

Solve the following LPP graphically: Maximize: \[ Z = 2x + 3y \] Subject to: \[ \begin{aligned} x + 4y &\leq 8 \quad \text{(1)} \\ 2x + 3y &\leq 12 \quad \text{(2)} \\ 3x + y &\leq 9 \quad \text{(3)} \\ x &\geq 0,\quad y \geq 0 \quad \text{(non-negativity constraints)} \end{aligned} \]

Answer 1

Step 1: Convert inequalities to equations

• Line 1: \( x + 4y = 8 \)
• Line 2: \( 2x + 3y = 12 \)
• Line 3: \( 3x + y = 9 \)

Step 2: Find Intercepts

• Line 1: \( x = 0 \Rightarrow y = 2 \), \( y = 0 \Rightarrow x = 8 \)
• Line 2: \( x = 0 \Rightarrow y = 4 \), \( y = 0 \Rightarrow x = 6 \)
• Line 3: \( x = 0 \Rightarrow y = 9 \), \( y = 0 \Rightarrow x = 3 \)

Step 3: Graph and Feasible Region

Plot all three lines and shade the region that satisfies all constraints including \( x \geq 0 \) and \( y \geq 0 \).

Step 4: Find Corner Points (Intersections)

1. Line 1 & Line 2:
   Solve: \[ x + 4y = 8 \quad \text{(i)} \\ 2x + 3y = 12 \quad \text{(ii)} \] Multiply (i) by 2: \[ 2x + 8y = 16 \] Subtract from (ii): \[ (2x + 3y) - (2x + 8y) = 12 - 16 \Rightarrow -5y = -4 \Rightarrow y = \frac{4}{5} \Rightarrow x = 8 - 4 \cdot \frac{4}{5} = \frac{24}{5} \] So point A: \( \left( \frac{24}{5}, \frac{4}{5} \right) \)
2. Line 2 & Line 3: 
   \[ 2x + 3y = 12,\quad 3x + y = 9 \Rightarrow y = 9 - 3x \] Substitute: \[ 2x + 3(9 - 3x) = 12 \Rightarrow 2x + 27 - 9x = 12 \Rightarrow -7x = -15 \Rightarrow x = \frac{15}{7} \Rightarrow y = 9 - 3 \cdot \frac{15}{7} = \frac{18}{7} \] So point B: \( \left( \frac{15}{7}, \frac{18}{7} \right) \)
3. Line 1 & Line 3: 
   \[ x + 4y = 8,\quad 3x + y = 9 \Rightarrow y = 9 - 3x \] Substitute: \[ x + 4(9 - 3x) = 8 \Rightarrow x + 36 - 12x = 8 \Rightarrow -11x = -28 \Rightarrow x = \frac{28}{11} \Rightarrow y = 9 - 3 \cdot \frac{28}{11} = \frac{15}{11} \] So point C: \( \left( \frac{28}{11}, \frac{15}{11} \right) \)

Step 5: Evaluate \( Z = 2x + 3y \) at Each Corner Point

Point	\( Z = 2x + 3y \)
(0, 0)	0
\( \left( \frac{24}{5}, \frac{4}{5} \right) \)	\[ Z = 2 \cdot \frac{24}{5} + 3 \cdot \frac{4}{5} = \frac{48 + 12}{5} = \frac{60}{5} = \boxed{12} \]
\( \left( \frac{15}{7}, \frac{18}{7} \right) \)	\[ Z = \frac{30}{7} + \frac{54}{7} = \frac{84}{7} = \boxed{12} \]
\( \left( \frac{28}{11}, \frac{15}{11} \right) \)	\[ Z = \frac{56}{11} + \frac{45}{11} = \frac{101}{11} \approx 9.18 \]

✅ Final Answer:

Maximum value of \( Z = 2x + 3y \) is \[ \boxed{12} \] at two corner points: \[ \left( \frac{24}{5}, \frac{4}{5} \right) \quad \text{and} \quad \left( \frac{15}{7}, \frac{18}{7} \right) \]