Question

A manufacturer makes two types of toys A and B. Three machines are needed for production with the following time constraints (in minutes): \[ \begin{array}{|c|c|c|} \hline \text{Machine} & \text{Toy A} & \text{Toy B} \\ \hline M1 & 12 & 6 \\ M2 & 18 & 0 \\ M3 & 6 & 9 \\ \hline \end{array} \] Each machine is available for 6 hours = 360 minutes. Profit on A = Rupee 20, on B = Rupee 30.
Formulate and solve the LPP graphically.

Hint:

Always graph constraints with proper scaling. The feasible region is where all constraints overlap. Evaluate the objective function at each corner point to find the maximum or minimum.

Answer 1

Step 1: Let the variables be Let \( x \) = number of Toy A units produced
Let \( y \) = number of Toy B units produced 
Step 2: Write the Objective Function We want to maximize profit: \[ Z = 20x + 30y \] 
Step 3: Translate constraints from machine limits

• M1: \( 12x + 6y \leq 360 \Rightarrow 2x + y \leq 60 \)
• M2: \( 18x \leq 360 \Rightarrow x \leq 20 \)
• M3: \( 6x + 9y \leq 360 \Rightarrow 2x + 3y \leq 120 \)
• Non-negativity: \( x \geq 0, \quad y \geq 0 \)

Step 4: Draw the Feasible Region 

Step 5: Find Corner Points of Feasible Region We solve the equations of intersecting lines to find vertices of the region: 
(i) Intersection of \( 2x + y = 60 \) and \( 2x + 3y = 120 \): \[ \text{From } 2x + y = 60 \Rightarrow y = 60 - 2x \\ \text{Substitute in } 2x + 3y = 120: \\ 2x + 3(60 - 2x) = 120 \Rightarrow 2x + 180 - 6x = 120 \Rightarrow -4x = -60 \Rightarrow x = 15 \\ \Rightarrow y = 60 - 2(15) = 30 \\ \Rightarrow \text{Point: } (15, 30) \] (ii) Intersection of \( x = 20 \) and \( 2x + y = 60 \): \[ 2(20) + y = 60 \Rightarrow y = 20 \Rightarrow \text{Point: } (20, 20) \] (iii) Intersection of \( x = 0 \) and \( 2x + 3y = 120 \): \[ 0 + 3y = 120 \Rightarrow y = 40 \Rightarrow \text{Point: } (0, 40) \] 
Step 6: Evaluate Objective Function at Each Corner Point \[ Z(0, 40) = 20(0) + 30(40) = 1200 \\ Z(15, 30) = 20(15) + 30(30) = 300 + 900 = 1200 \\ Z(20, 20) = 20(20) + 30(20) = 400 + 600 = 1000 \] 
Step 7: Select Optimal Value \( \text{Maximum profit }\) \(Z =  1200 \text{ at points } (0, 40) \text{ and } (15, 30) \)
Final Answer: 
\[ \boxed{ \text{Maximum profit is }  1200 \text{ when } (x, y) = (0, 40) \text{ or } (15, 30) } \]