Constrained Extrema on an Ellipsoid (Lagrange Multipliers): 04-BS-1, December 2018, Q3

Question 3: Constrained Extrema on an Ellipsoid (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 $f=3x+2y^2+z$; constraint $g=3x^2+y^2+z^2-1=0$ (a compact surface, so global extrema exist).

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

Approach. Set $\nabla f=\lambda\nabla g$ componentwise, solve the resulting system (branching on whether $y=0$), keep only the branches that also satisfy the constraint, then compare all the resulting $f$-values.

Lagrange equations. $\nabla f=(3,4y,1)$, $\nabla g=(6x,2y,2z)$, giving $$3=6\lambda x,\qquad 4y=2\lambda y,\qquad 1=2\lambda z.$$ The middle equation factors as $y(4-2\lambda)=0$, so either $y=0$ or $\lambda=2$.
Branch $y=0$. Then $x=\dfrac1{2\lambda}=z$. Substituting into $3x^2+z^2=1$ (with $y=0$): $3x^2+x^2=1\Rightarrow x^2=\tfrac14\Rightarrow x=\pm\tfrac12$, so $\lambda=\pm1$ and the two critical points are $\left(\tfrac12,0,\tfrac12\right)$, $\left(-\tfrac12,0,-\tfrac12\right)$, giving $f=2$ and $f=-2$ respectively.
Branch $\lambda=2$. Then $x=\dfrac1{4}$, $z=\dfrac1{4}$ (from $3=12x$ and $1=4z$). Constraint: $3\left(\tfrac1{16}\right)+y^2+\tfrac1{16}=1\Rightarrow y^2=\tfrac34\Rightarrow y=\pm\dfrac{\sqrt3}2$, giving the two points $\left(\tfrac14,\pm\tfrac{\sqrt3}2,\tfrac14\right)$, both with $$f=3\!\left(\tfrac14\right)+2\!\left(\tfrac34\right)+\tfrac14=\tfrac34+\tfrac32+\tfrac14=\tfrac52.$$
Compare all critical values. The four critical points give $f\in\{2,\,-2,\,\tfrac52,\,\tfrac52\}$. Since the ellipsoid is compact, the extrema exist among these: $$f_{\max}=\tfrac52\ \text{at}\ \left(\tfrac14,\pm\tfrac{\sqrt3}2,\tfrac14\right),\qquad f_{\min}=-2\ \text{at}\ \left(-\tfrac12,0,-\tfrac12\right).$$ (The point $\left(\tfrac12,0,\tfrac12\right)$ with $f=2$ is a third critical point but neither the global max nor min.)

$$f_{\max}=\boxed{\tfrac52}\qquad\qquad f_{\min}=\boxed{-2}$$

Quantity	Result
Critical values	$-2,\ 2,\ \tfrac52,\ \tfrac52$
Maximum	$\tfrac52$ at $\left(\tfrac14,\pm\tfrac{\sqrt3}2,\tfrac14\right)$
Minimum	$-2$ at $\left(-\tfrac12,0,-\tfrac12\right)$

Concept & Theory

Topic. Lagrange multipliers for constrained optimization on a compact surface.

Key relations. At a constrained extremum, $\nabla f=\lambda\nabla g$ for some scalar $\lambda$; combined with $g=0$ this gives a system for the critical points. On a compact constraint surface (closed and bounded, like an ellipsoid), the global max/min are guaranteed to occur among the finitely many critical points.

Why this works. $\nabla f$ must be parallel to $\nabla g$ at a constrained extremum because any component of $\nabla f$ tangent to the constraint surface would allow $f$ to increase or decrease by moving along the surface, contradicting extremality.

Common pitfall. Dropping the $y=0$ branch by dividing the middle equation by $y$ without checking $y=0$ separately; forgetting to check all critical values before naming the max/min (the middle value $f=2$ here is a genuine critical point but not an extremum).

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

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

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