Class 12 Linear Programming · Case Study4 Markslinear_programming_models
Class 12 Maths · Linear Programming · Case Study #73
A feasible polygon has boundary vertices O(0,0),A(0,6),B(4,10),C(6,6),D(6,4),E(5,0). Answer (i),(ii), and either (iii)(a) or (iii)(b).
Part (1) 1 Marks
Is the region bounded?
Answer: Yes.
- All its points lie in the rectangle 0≤x≤6,0≤y≤10.
- Thus the polygon is bounded.
Part (2) 1 Marks
How many vertices does it have?
Answer: 6.
- List O,A,B,C,D,E, including the origin.
- There are six vertices.
Part (3) 2 Marks
If Z=ax+by has maxima at both B and C, derive the relation between a,b.
Answer: a=2b; for these points to be maxima, b≥0.
- Equal values give 4a+10b=6a+6b.
- Hence 2a=4b and a=2b. Then their common value 18b dominates all other vertex values precisely when b≥0.
- Thus a=2b, with b≥0 for the stated common maxima.
Part (4) 2 Marks
Find min Z=2x−5y.
Answer: −42 at B(4,10).
- Evaluate all six vertices by the corner-point theorem.
- The values at O,A,B,C,D,E are 0,−30,−42,−18,−8,10.
- Minimum is −42 at B(4,10).