UPSC 2016 Maths Optional Paper 1 Q3a — Step-by-Step Solution

20 marks · Section A

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

Question

Find the maximum and minimum values of $x^2+y^2+z^2$ subject to the conditions $\dfrac{x^2}{4}+\dfrac{y^2}{5}+\dfrac{z^2}{25}=1$ and $x+y-z=0$ .

Technique

Lagrange multipliers with two constraints reduce (for $r^2$ on a central plane section of a quadric) to the secular equation $\sum \ell_i^2 a_i/(a_i-r^2)=0$ ; its roots are the extreme squared distances.

Solution

We maximize/minimize $r^2=x^2+y^2+z^2$ on the ellipse cut from the ellipsoid by the plane.

Step 1 — Lagrange conditions

With multipliers $\lambda$ (ellipsoid) and $\mu$ (plane), $\nabla(x^2+y^2+z^2)=\lambda\nabla g+\mu\nabla h$ gives

2x=\lambda\tfrac{2x}{4}+\mu,\quad 2y=\lambda\tfrac{2y}{5}+\mu,\quad 2z=\lambda\tfrac{2z}{25}-\mu.

Write the ellipsoid denominators as $a_1=4,a_2=5,a_3=25$ . The stationary $r^2$ -values are the roots of the secular equation

\sum_{i}\frac{\ell_i^2\,a_i}{a_i-r^2}=0,

where $(\ell_1,\ell_2,\ell_3)=(1,1,-1)$ are the plane’s coefficients (standard two-constraint result for “distance $^2$ extremized on a central plane section of a quadric”).

Step 2 — Form and solve the secular equation

\frac{4}{4-r^2}+\frac{5}{5-r^2}+\frac{25}{25-r^2}=0.

Clear denominators. Multiplying out gives a quadratic in $s=r^2$ :

17s^2-245s+750=0\ \Longrightarrow\ s=\frac{245\pm\sqrt{245^2-4\cdot17\cdot750}}{34}=\frac{245\pm115}{34}\ \Longrightarrow\ s=10\ \text{or}\ s=\tfrac{75}{17}.

Step 3 — Identify max and min

Answer

\boxed{\;\max\big(x^2+y^2+z^2\big)=10,\qquad \min\big(x^2+y^2+z^2\big)=\frac{75}{17}\approx4.412.\;}