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UPSC 2021 Maths Optional Paper 2 Q3c — Step-by-Step Solution

15 marks · Section A

Duality · Linear Programming · asked 4× in 13 yrs · Read the full method →

Question

Convert into dual:

subject to

$x_1,x_2\ge 0$, $x_3$ unrestricted.

Technique

Apply primal-dual conversion rules; track sign restrictions based on Min/Max and $\le/\ge/=$ direction.

Solution

Step 1 — Convert to standard primal form (Min with $\ge$ constraints)

For minimisation, the standard dual form uses $\ge$ constraints.

Convert:

• $3x_1-x_2+2x_3\le 7\Rightarrow -3x_1+x_2-2x_3\ge -7$.
• $2x_1-4x_2\ge 12$ — already $\ge$.
• $-4x_1+3x_2+8x_3=10$ — equality. Split into two $\ge$: $-4x_1+3x_2+8x_3\ge 10$ and $4x_1-3x_2-8x_3\ge -10$ (or treat equality as unrestricted-sign dual variable).

Better approach: use the direct rules for primal–dual:

Step 2 — Apply rules

Primal (Min):

• Constraint 1: $3x_1-x_2+2x_3\le 7$ → dual variable $y_1\le 0$.
• Constraint 2: $2x_1-4x_2\ge 12$ → dual variable $y_2\ge 0$.
• Constraint 3: $-4x_1+3x_2+8x_3=10$ → dual variable $y_3$ unrestricted.

Primal variables:

• $x_1\ge 0$ → dual constraint $\le c_1=1$.
• $x_2\ge 0$ → dual constraint $\le c_2=-3$.
• $x_3$ unrestricted → dual constraint $=c_3=-2$.

Step 3 — Build dual

Dual: Maximize $W=7y_1+12y_2+10y_3$

subject to:

• (for $x_1$): $3y_1+2y_2-4y_3\le 1$.
• (for $x_2$): $-y_1-4y_2+3y_3\le-3$.
• (for $x_3$): $2y_1+0\cdot y_2+8y_3=-2$, i.e., $2y_1+8y_3=-2$, equivalently $y_1+4y_3=-1$.

with $y_1\le 0$, $y_2\ge 0$, $y_3$ unrestricted.

Answer