Question 8 of 8: Lagrange Multipliers — Minimum on an Ellipsoid
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2018 — 04-BS-1 Mathematics. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved calculator allowed). Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Given. Linear objective $F(x,y,z)=x-y+2z$; ellipsoidal constraint $g(x,y,z)=x^2+3y^2+2z^2=5$.
Find. The minimum value of $F$ on the constraint surface.
Approach. Since the ellipsoid is compact, $F$ attains both a maximum and minimum on it. Apply Lagrange multipliers: $\nabla F=\lambda\nabla g$, solve for the two critical points in terms of $\lambda$, substitute into the constraint to fix $\lambda$, then evaluate $F$ at both to identify the minimum.
Substitute into the constraint.
$$x^2+3y^2+2z^2=\frac1{4\lambda^2}+\frac1{12\lambda^2}+\frac1{2\lambda^2}=\frac{3+1+6}{12\lambda^2}=\frac{10}{12\lambda^2}=\frac5{6\lambda^2}=5\ \Longrightarrow\ \lambda^2=\frac16,\ \ \lambda=\pm\frac1{\sqrt6}.$$
Evaluate $F$ at each critical point.
$$F=x-y+2z=\frac1{2\lambda}+\frac1{6\lambda}+\frac1\lambda=\frac{3+1+6}{6\lambda}=\frac{10}{6\lambda}=\frac5{3\lambda}.$$
For $\lambda=\tfrac1{\sqrt6}$: $F=\dfrac{5\sqrt6}3$. For $\lambda=-\tfrac1{\sqrt6}$: $F=-\dfrac{5\sqrt6}3$.
Identify the minimum. The two critical values are $\pm\dfrac{5\sqrt6}3$; the minimum is the negative one, attained at $\lambda=-1/\sqrt6$, i.e. $(x,y,z)=\left(-\tfrac{\sqrt6}2,\ \tfrac{\sqrt6}6,\ -\tfrac{\sqrt6}2\right)$.
Cross-check (Cauchy–Schwarz): writing $u=x,\ v=\sqrt3y,\ w=\sqrt2z$ turns the constraint into $u^2+v^2+w^2=5$ and $F=u-\tfrac1{\sqrt3}v+\sqrt2w$, whose extrema on the sphere are $\pm\sqrt5\cdot\sqrt{1+\tfrac13+2}=\pm\sqrt5\cdot\sqrt{10/3}=\pm\tfrac{5\sqrt6}3$ — matches exactly.
