GATE ME 2020 Set 1 — Question 57

1. Define Variables and Objective Functions:
Let $x_A$ and $x_B$ be the quantities (in kg) of products A and B, respectively.For both Shveta and Ashok, the selling prices are:
Product A: Rs. 2,000 per kg
Product B: Rs. 3,000 per kg2. Formulate Shveta's Linear Programming Problem:
Objective: Maximize Revenue $R_S = 2000 x_A + 3000 x_B$Constraints for Shveta:
(i) Daily production of A has to be at least as much as B: $x_A \ge x_B$
(ii) Upper limit for production of B is 10 kg per day: $x_B \le 10$
(iii) State regulations restrict Shveta's production of A to less than 20 kg per day: $x_A < 20$
(iv) Non-negativity constraints: $x_A \ge 0, x_B \ge 0$To find the optimal revenue, we consider the boundary of the feasible region. For $x_A < 20$, the maximum value of $x_A$ approaches 20. So, we evaluate the objective function at the vertices of the feasible region defined by $x_A \ge x_B$, $x_B \le 10$, $x_A \le 20$, $x_A \ge 0, x_B \ge 0$.Vertices of Shveta's feasible region:

• $(0,0) \implies R_S = 2000(0) + 3000(0) = 0$
• $(10,10)$ (from $x_A = x_B$ and $x_B = 10$) $\implies R_S = 2000(10) + 3000(10) = 20000 + 30000 = 50000$
• $(20,10)$ (from $x_A = 20$ and $x_B = 10$) $\implies R_S = 2000(20) + 3000(10) = 40000 + 30000 = 70000$
• $(20,0)$ (from $x_A = 20$ and $x_B = 0$) $\implies R_S = 2000(20) + 3000(0) = 40000$

The maximum revenue for Shveta is Rs. 70,000.3. Formulate Ashok's Linear Programming Problem:
Objective: Maximize Revenue $R_A = 2000 y_A + 3000 y_B$Constraints for Ashok:
(i) Daily production of A has to be at least as much as B: $y_A \ge y_B$
(ii) Upper limit for production of B is 10 kg per day: $y_B \le 10$
(iii) State regulations restrict Ashok's production of A to less than 15 kg per day: $y_A < 15$
(iv) Non-negativity constraints: $y_A \ge 0, y_B \ge 0$Similar to Shveta's problem, we consider $y_A \le 15$ for finding the supremum of the revenue.Vertices of Ashok's feasible region:

• $(0,0) \implies R_A = 2000(0) + 3000(0) = 0$
• $(10,10)$ (from $y_A = y_B$ and $y_B = 10$) $\implies R_A = 2000(10) + 3000(10) = 20000 + 30000 = 50000$
• $(15,10)$ (from $y_A = 15$ and $y_B = 10$) $\implies R_A = 2000(15) + 3000(10) = 30000 + 30000 = 60000$
• $(15,0)$ (from $y_A = 15$ and $y_B = 0$) $\implies R_A = 2000(15) + 3000(0) = 30000$

The maximum revenue for Ashok is Rs. 60,000.4. Calculate the Absolute Difference in Optimal Revenue:
Absolute difference $= |R_S - R_A| = |70000 - 60000| = 10000$ Rupees.5. Convert to Thousand Rupees and Round Off:
$10000 \text{ Rupees} = \frac{10000}{1000} \text{ thousand Rupees} = 10 \text{ thousand Rupees}$.
Rounding off to 2 decimal places, the value is $10.00$.The final answer is $\boxed{10.00}$.