Q5. A tailor makes shirts and trousers. A shirt needs 2 m of cloth and 2 hours, a trouser 3 m of cloth and 1 hour. She has 24 m of cloth and 16 hours, and earns Rs 30 on a shirt and Rs 40 on a trouser. Formulate the problem and find the most profitable plan.
With x shirts and y trousers: maximise Z = 30x + 40y subject to 2x + 3y ≤ 24, 2x + y ≤ 16, x, y ≥ 0. The lines meet at (6, 4). The corners (0, 0), (8, 0), (6, 4) and (0, 8) give Z = 0, 240, 340 and 320, so she should make 6 shirts and 4 trousers, for Rs 340.
Q6. A diet must contain at least 8 units of vitamin and 7 units of mineral. Food A (Rs 4 a unit) gives 2 units of vitamin and 1 of mineral; food B (Rs 5 a unit) gives 1 of vitamin and 2 of mineral. Find the cheapest diet.
Minimise Z = 4x + 5y subject to 2x + y ≥ 8, x + 2y ≥ 7, x, y ≥ 0. The region is unbounded, with corners (0, 8), (3, 2) and (7, 0), where Z is 40, 22 and 28. The half plane 4x + 5y < 22 has no point in the region, because 4x + 5y = (2x + y) + 2(x + 2y) ≥ 8 + 14 = 22, so the minimum is Rs 22: 3 units of A and 2 of B.
Q7. For the constraints 2x − y ≥ −5, 3x + y ≥ 3, 2x − 3y ≤ 12 and x, y ≥ 0 of NCERT Example 4 (pages 402–403), can Z = −50x + 20y have the maximum value 100 at (0, 5)?
Check whether −50x + 20y > 100 meets the feasible region. The constraint 2x − y ≥ −5 gives y ≤ 2x + 5 on the whole region, so there 20y ≤ 40x + 100 and Z ≤ −10x + 100 ≤ 100. The half plane has no point in the region, so 100 is the maximum.
Q8. What does the dealer’s problem become if he can store 70 pieces instead of 60?
The constraint becomes x + y ≤ 70. With 5x + y ≤ 100, the lines meet at (7.5, 62.5). The corners (0, 0), (20, 0), (7.5, 62.5) and (0, 70) give Z = 0, 5000, 6562.5 and 5250, so the best plan moves to 7.5 tables and 62.5 chairs. That plan is not in whole numbers. Among whole-number plans the best is 8 tables and 60 chairs (5 × 8 + 60 = 100, 68 ≤ 70), for Rs 6500. A graph alone does not guarantee whole numbers.
Q9. Why must the optimal value of a linear objective function on a bounded feasible region occur at a corner?
Assume the feasible region is nonempty and a, b are not both zero. The lines ax + by = k are parallel. Slide such a line across the region: the last point it touches as k grows (or falls) is a corner, or a whole edge containing two corners. If a = b = 0, Z = 0 everywhere, so every feasible point, including every corner, is optimal. NCERT states the corner result as Theorems 1 and 2 without proof (page 398).