Constrained Extrema on an Ellipsoid via Lagrange Multipliers: 04-BS-1, December 2016, Q3

Question 3: Constrained Extrema on an Ellipsoid via Lagrange Multipliers 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Objective $f(x,y,z)=4x+y^2+2z^2$ on the compact constraint surface $g(x,y,z)=x^2+3y^2+z^2-2=0$ (an ellipsoid, so a global max and min are guaranteed to exist).

Find. The global maximum and minimum values of $f$ on the ellipsoid.

Approach. Solve $\nabla f=\lambda\nabla g$ together with the constraint; because two of the three component equations factor ($y(\cdots)=0$, $z(\cdots)=0$), split into cases by which variables vanish, then compare $f$ across all resulting critical points.

Set up the Lagrange conditions. $\nabla f=(4,2y,4z)$, $\nabla g=(2x,6y,2z)$, so $$4=2\lambda x,\qquad 2y=6\lambda y,\qquad 4z=2\lambda z.$$ The last two factor as $y(1-3\lambda)=0$ and $z(2-\lambda)=0$, giving four cases.
Case A: $y=0,\,z=0$. Constraint gives $x^2=2\Rightarrow x=\pm\sqrt2$. Then $f=4x=\pm4\sqrt2\approx\pm5.657$.
Case B: $y=0,\,\lambda=2$. From $4=2\lambda x$, $x=1$. Constraint: $1+z^2=2\Rightarrow z=\pm1$. Then $$f=4(1)+0+2(1)=6.$$
Case C: $z=0,\,\lambda=\tfrac13$. From $4=2\lambda x$, $x=6$. Constraint: $36+3y^2=2\Rightarrow y^2=-\tfrac{34}{3}$, no real solution — this case contributes nothing.
Case D: $\lambda=\tfrac13$ and $\lambda=2$ simultaneously. Impossible (a single $\lambda$ cannot equal both), so no critical point has both $y\ne0$ and $z\ne0$.
Compare all valid critical values. The candidates are $f=4\sqrt2\approx5.657$, $f=-4\sqrt2\approx-5.657$, and $f=6$. Since the ellipsoid is compact and $f$ continuous, the global extrema must occur among these: $$f_{\max}=\boxed{6}\text{ at }(1,0,\pm1),\qquad f_{\min}=\boxed{-4\sqrt2}\text{ at }(-\sqrt2,0,0)$$

Critical point	$f$ value
$(\sqrt2,0,0)$	$4\sqrt2\approx5.657$
$(-\sqrt2,0,0)$	$-4\sqrt2\approx-5.657$ (global min)
$(1,0,\pm1)$	$6$ (global max)

Concept & Theory

Topic. Constrained global extrema on a compact surface via Lagrange multipliers.

Key relations. At an extremum of $f$ subject to $g=0$, $\nabla f=\lambda\nabla g$. On a compact (closed and bounded) constraint set, continuous $f$ attains both a global max and min, so it suffices to evaluate $f$ at every critical point produced by the Lagrange system and take the largest/smallest.

Why this works. Two of the three gradient-matching equations here are already factored ($y(1-3\lambda)=0$, $z(2-\lambda)=0$) because $y^2,z^2$ appear in both $f$ and $g$ with only even powers — this structure is what produces clean either/or branches instead of a fully coupled nonlinear system.

Common pitfall. Stopping after finding only Case A or only Case B and reporting its value as the answer without checking every branch of the case split; missing that Case C is vacuous (no real solution) and mistakenly trying to force a value from it; forgetting that the ellipsoid's compactness is what justifies comparing finitely many critical points rather than needing a second-derivative test.

Source. Stewart, Calculus: Early Transcendentals, Ch. 14 (§14.8 Lagrange Multipliers).

Question 3 of 8: Constrained Extrema on an Ellipsoid via Lagrange Multipliers

Question 3: Constrained Extrema on an Ellipsoid via Lagrange Multipliers 20 marks