UPSC 2019 Maths Optional Paper 2 Q4a — Step-by-Step Solution

15 marks · Section A

Lagrange's method of multipliers (constrained extrema) · Calculus · asked 8× in 13 yrs · Read the full method →

Question

Find the maximum value of $f(x,y,z)=x^2y^2z^2$ subject to the subsidiary condition $x^2+y^2+z^2=c^2,\ (x,y,z>0)$ .

Lagrange multipliers on a symmetric objective; the symmetry forces $x=y=z$ . Cross-check by AM–GM on the squares.

Maximize $f=x^2y^2z^2$ subject to $g=x^2+y^2+z^2-c^2=0$ . Set $\nabla f=\lambda\nabla g$ :

2xy^2z^2=\lambda\,2x,\qquad 2x^2yz^2=\lambda\,2y,\qquad 2x^2y^2z=\lambda\,2z.

Since $x,y,z>0$ , divide the first by $2x$ , the second by $2y$ , the third by $2z$ :

y^2z^2=\lambda,\qquad x^2z^2=\lambda,\qquad x^2y^2=\lambda.

From $y^2z^2=x^2z^2$ (and $z>0$ ): $y^2=x^2\Rightarrow y=x$ . From $x^2z^2=x^2y^2$ (and $x>0$ ): $z^2=y^2\Rightarrow z=y$ . Hence

x=y=z.

Impose the constraint: $3x^2=c^2\Rightarrow x^2=\dfrac{c^2}{3}$ , so

x=y=z=\frac{c}{\sqrt3}.

f_{\max}=x^2y^2z^2=(x^2)^3=\left(\frac{c^2}{3}\right)^3=\frac{c^6}{27}.

\boxed{\;\max\,(x^2y^2z^2)=\frac{c^6}{27},\quad\text{at } x=y=z=\frac{c}{\sqrt3}.\;}