← 2015 Paper 2
UPSC 2015 Maths Optional Paper 2 Q3c-ii — Step-by-Step Solution
10 marks · Section A
LPP: standard form; basic, basic feasible, optimal solutions · Linear Programming · asked 6× in 13 yrs · Read the full method →
Question
Without solving the problem, show that it has an optimal solution. Which of the basic feasible solution(s) is/are optimal?
Technique
Substitute equality constraints to reduce Z to a single variable x3; observe Z=20−7x3 is bounded above (since x3≥0); by LPP fundamental theorem, optimum is attained; check each BFS.
Solution
LPP (recap):
Maximize Z=x1+2x2−3x3+4x4,x1+x2+2x3+3x4=12,x2+2x3+x4=8,xi≥0.
Step 1 — Existence of optimal solution
Theorem (LPP fundamental). A linear program with non-empty feasible region and bounded objective on the feasible region attains its optimum at an extreme point (= basic feasible solution).
Non-empty feasible region: From Q3(c)(i), we have at least one BFS — e.g., (4,8,0,0) — so the feasible region is non-empty.
Boundedness of Z above: Let’s show Z is bounded above on the feasible region.
From the second constraint x2+2x3+x4=8, we have x4=8−x2−2x3.
Substitute into the first: x1+x2+2x3+3(8−x2−2x3)=12, so x1=12−x2−2x3−24+3x2+6x3=−12+2x2+4x3.
So x1=2x2+4x3−12 and x4=8−x2−x4… wait, x4=8−x2−2x3.
Substitute into Z:
Z=x1+2x2−3x3+4x4=(2x2+4x3−12)+2x2−3x3+4(8−x2−2x3)
=2x2+4x3−12+2x2−3x3+32−4x2−8x3
=(2+2−4)x2+(4−3−8)x3+(32−12)
=0⋅x2−7x3+20.
So Z=20−7x3 on the feasible region (with x1≥0,x4≥0 as additional constraints).
Since x3≥0, Z=20−7x3≤20, with equality iff x3=0. So Z is bounded above by 20.
By the fundamental theorem, the LPP has an optimal BFS, attaining Z=20.
Step 2 — Identify the optimal BFS
Z=20−7x3 is maximized when x3=0. Looking at the 4 BFS from Q3(c)(i):
| BFS | x3 | Z=20−7x3 |
|---|
| (4,8,0,0) | 0 | 20 |
| (4,0,4,0) | 4 | 20−28=−8 |
| (0,6,0,2) | 0 | 20 |
| (0,0,3,2) | 3 | 20−21=−1 |
Two BFS attain the maximum Z=20: (4,8,0,0) and (0,6,0,2).
Answer
Zmax=20, attained at BFS (4,8,0,0) and (0,6,0,2).