Constrained Minimum via Lagrange Multipliers: 04-BS-1, December 2013, Q3

Question 3: Constrained Minimum 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 $F(x,y,z)=x-y+2z$; constraint $g(x,y,z)=x^2+3y^2+2z^2-5=0$ (a bounded ellipsoid).

Find. The minimum value of $F$ on the ellipsoid.

Approach. Since the constraint set is a compact ellipsoid, $F$ attains both a max and min on it. Use Lagrange multipliers: solve $\nabla F=\lambda\nabla g$ together with the constraint, giving two critical points (one max, one min).

Set up the Lagrange conditions. $\nabla F=(1,-1,2)$, $\nabla g=(2x,6y,4z)$, so $$1=2\lambda x,\qquad -1=6\lambda y,\qquad 2=4\lambda z.$$ Solving each for the variable: $x=\dfrac{1}{2\lambda},\ y=-\dfrac{1}{6\lambda},\ z=\dfrac{1}{2\lambda}$.
Substitute into the constraint. $$x^2+3y^2+2z^2=\frac{1}{4\lambda^2}+\frac{3}{36\lambda^2}+\frac{2}{4\lambda^2}=\frac{1}{4\lambda^2}+\frac{1}{12\lambda^2}+\frac{1}{2\lambda^2}=\frac{10}{12\lambda^2}=\frac{5}{6\lambda^2}=5.$$ $$\Rightarrow\quad \lambda^2=\frac16 \;\Rightarrow\; \lambda=\pm\frac{1}{\sqrt6}.$$
Evaluate $F$ at both critical points. $$F=x-y+2z=\frac{1}{2\lambda}+\frac{1}{6\lambda}+\frac{1}{\lambda}=\frac{3+1+6}{6\lambda}=\frac{10}{6\lambda}=\frac{5}{3\lambda}.$$ For $\lambda=+\tfrac1{\sqrt6}$: $F=\tfrac{5\sqrt6}{3}$ (maximum). For $\lambda=-\tfrac1{\sqrt6}$: $F=-\tfrac{5\sqrt6}{3}$ (minimum).

$$F_{\min}=\boxed{-\dfrac{5\sqrt6}{3}}\approx -4.082 \quad\text{at}\quad (x,y,z)=\left(-\tfrac{\sqrt6}{2},\ \tfrac{\sqrt6}{6},\ -\tfrac{\sqrt6}{2}\right)$$

Quantity	Result
$\lambda$ at the minimum	$-1/\sqrt6$
$(x,y,z)$ at the minimum	$(-\sqrt6/2,\ \sqrt6/6,\ -\sqrt6/2)$
$F_{\min}$	$-\dfrac{5\sqrt6}{3}\approx-4.082$

Concept & Theory

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

Key relations. At a constrained extremum, $\nabla F=\lambda\nabla g$ for some scalar $\lambda$ (the gradients are parallel because moving along the constraint surface produces no first-order change in $F$ unless $\nabla F$ has a component normal to the surface, which is exactly $\lambda\nabla g$).

Why this works. The ellipsoid $x^2+3y^2+2z^2=5$ is closed and bounded (compact), so by the extreme value theorem $F$ (continuous) attains a global max and min on it; the Lagrange system has exactly two solutions (the two signs of $\lambda$), which must be exactly those two extrema.

Common pitfall. Only reporting one critical point and not checking which of the two candidate values is the minimum vs. maximum; also arithmetic slips combining the fractions $\tfrac{1}{4\lambda^2}+\tfrac{1}{12\lambda^2}+\tfrac{1}{2\lambda^2}$ over a common denominator.

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

Question 3 of 8: Constrained Minimum via Lagrange Multipliers

Question 3: Constrained Minimum via Lagrange Multipliers (20 marks)