UPSC 2023 Maths Optional Paper 2 Q2b — 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

Using the method of Lagrange’s multipliers, find the minimum and maximum distances of the point $P(2,6,3)$ from the sphere $x^2+y^2+z^2=4$ .

Technique

Lagrange multipliers; the critical points turn out to be the two intersections of the line $OP$ with the sphere.

Solution

Strategy. Minimise/maximise the squared distance $f(x,y,z)=(x-2)^{2}+(y-6)^{2}+(z-3)^{2}$ subject to $g(x,y,z)=x^{2}+y^{2}+z^{2}-4=0$ . Squaring is monotone on non-negative reals, so extremising $f$ extremises the actual distance.

Step 1 — Lagrange system.

$\nabla f=\lambda\nabla g$ gives

2(x-2)=2\lambda x,\quad 2(y-6)=2\lambda y,\quad 2(z-3)=2\lambda z,

i.e.

(1-\lambda)x=2,\quad(1-\lambda)y=6,\quad(1-\lambda)z=3. \tag{$\ast$}

Note $\lambda\ne 1$ (otherwise the LHS of $(\ast)$ vanishes but the RHS doesn’t), so

x=\frac{2}{1-\lambda},\quad y=\frac{6}{1-\lambda},\quad z=\frac{3}{1-\lambda}.

Step 2 — Apply the constraint.

x^{2}+y^{2}+z^{2}=\frac{4+36+9}{(1-\lambda)^{2}}=\frac{49}{(1-\lambda)^{2}}=4

\Rightarrow\;(1-\lambda)^{2}=\frac{49}{4}\;\Rightarrow\;1-\lambda=\pm\frac{7}{2}.

Step 3 — Two critical points.

Case $1-\lambda=\tfrac{7}{2}$ : $\;(x,y,z)=\bigl(\tfrac{4}{7},\tfrac{12}{7},\tfrac{6}{7}\bigr)$ .
Case $1-\lambda=-\tfrac{7}{2}$ : $\;(x,y,z)=\bigl(-\tfrac{4}{7},-\tfrac{12}{7},-\tfrac{6}{7}\bigr)$ .

Step 4 — Evaluate $f$ at both.

Let $Q_{\pm}=\pm\bigl(\tfrac{4}{7},\tfrac{12}{7},\tfrac{6}{7}\bigr)=\pm\tfrac{2}{7}(2,6,3)=\pm\tfrac{2}{7}\mathbf{P}$ .

f(Q_{+})=\bigl|\mathbf{P}-\tfrac{2}{7}\mathbf{P}\bigr|^{2}=\bigl(\tfrac{5}{7}\bigr)^{2}|\mathbf{P}|^{2}=\tfrac{25}{49}\cdot 49=25,

f(Q_{-})=\bigl|\mathbf{P}+\tfrac{2}{7}\mathbf{P}\bigr|^{2}=\bigl(\tfrac{9}{7}\bigr)^{2}|\mathbf{P}|^{2}=\tfrac{81}{49}\cdot 49=81.

So the squared distances are $25$ (minimum) and $81$ (maximum).

Step 5 — Conclude.

Answer

\boxed{\;d_{\min}=5\text{ at }\bigl(\tfrac{4}{7},\tfrac{12}{7},\tfrac{6}{7}\bigr),\quad d_{\max}=9\text{ at }\bigl(-\tfrac{4}{7},-\tfrac{12}{7},-\tfrac{6}{7}\bigr).\;}