← 2014 Paper 2

UPSC 2014 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

Maxima and minima of multi-variable functions (analytic criteria) · Real Analysis · asked 5× in 14 yrs · Read the full method →

Question

Find the minimum value of x2+y2+z2x^{2}+y^{2}+z^{2} subject to the condition xyz=a3xyz=a^{3} by the method of Lagrange multipliers.

Technique

Standard 1-constraint Lagrange; symmetry of the objective (sum of squares) under sign flips and permutations.

Solution

Strategy. Standard Lagrange-multiplier setup; exploit symmetry to find critical points.

Step 1 — Lagrange equations

L=x2+y2+z2−λ(xyz−a3)L=x^{2}+y^{2}+z^{2}-\lambda(xyz-a^{3}).

∂L/∂x=2x−λyz=0,  ∂L/∂y=2y−λxz=0,  ∂L/∂z=2z−λxy=0.\partial L/\partial x=2x-\lambda yz=0,\;\partial L/\partial y=2y-\lambda xz=0,\;\partial L/\partial z=2z-\lambda xy=0.

Step 2 — Identify λ\lambda

Multiply first equation by xx: 2x2=λxyz=λa32x^{2}=\lambda xyz=\lambda a^{3}.

Similarly: 2y2=λa32y^{2}=\lambda a^{3}, 2z2=λa32z^{2}=\lambda a^{3}.

So x2=y2=z2x^{2}=y^{2}=z^{2}.

Step 3 — Determine signs from constraint

∣x∣=∣y∣=∣z∣|x|=|y|=|z|. From xyz=a3xyz=a^{3} (with a>0a>0 assumed, so RHS positive), either:

In either case, ∣x∣=∣y∣=∣z∣=t|x|=|y|=|z|=t for some t>0t>0, and ∣xyz∣=t3=∣a3∣=a3|xyz|=t^{3}=|a^{3}|=a^{3}, so t=at=a.

Step 4 — Compute the objective

x2+y2+z2=3t2=3a2x^{2}+y^{2}+z^{2}=3t^{2}=3a^{2}.

Answer

  min⁡(x2+y2+z2)=3a2,  attained at  (a,a,a) (and other sign patterns).  \boxed{\;\min(x^{2}+y^{2}+z^{2})=3a^{2},\;\text{attained at}\;(a,a,a)\text{ (and other sign patterns).}\;}
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