Constrained Max/Min via Lagrange Multipliers (Ellipsoid Constraint): 04-BS-1, December 2015, Q4

Question 4: Constrained Max/Min via Lagrange Multipliers (Ellipsoid Constraint) (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 $g(x,y,z)=x^2+y^2+4z^2-1=0$ (a compact ellipsoid).

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

Approach. The ellipsoid is closed and bounded, so $f$ (continuous) attains both a max and a min on it. Use Lagrange multipliers: $\nabla f=\lambda\nabla g$ together with the constraint gives exactly two critical points.

Set up the Lagrange conditions. $\nabla f=(1,1,-2)$, $\nabla g=(2x,2y,8z)$, so $$1=2\lambda x,\qquad 1=2\lambda y,\qquad -2=8\lambda z.$$ Solving each for the variable: $x=y=\dfrac{1}{2\lambda},\qquad z=-\dfrac{1}{4\lambda}.$
Substitute into the constraint. $$x^2+y^2+4z^2=\frac{2}{4\lambda^2}+\frac{4}{16\lambda^2}=\frac{1}{2\lambda^2}+\frac{1}{4\lambda^2}=\frac{3}{4\lambda^2}=1 \;\Rightarrow\; \lambda^2=\frac34 \;\Rightarrow\; \lambda=\pm\frac{\sqrt3}{2}.$$
Evaluate $f$ at both critical points. $$f=x+y-2z=\frac{1}{2\lambda}+\frac{1}{2\lambda}+\frac{1}{2\lambda}=\frac{3}{2\lambda}.$$ For $\lambda=+\tfrac{\sqrt3}{2}$: $f=\dfrac{3}{\sqrt3}=\sqrt3$ (maximum). For $\lambda=-\tfrac{\sqrt3}{2}$: $f=-\sqrt3$ (minimum).

$$f_{\max}=\boxed{\sqrt3}\ \text{at}\ \left(\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3},-\tfrac{1}{2\sqrt3}\right),\qquad f_{\min}=\boxed{-\sqrt3}\ \text{at}\ \left(-\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3},\tfrac{1}{2\sqrt3}\right)$$

Quantity	Result
$\lambda$ values	$\pm\sqrt3/2$
$f_{\max}$	$\sqrt3\approx1.732$
$f_{\min}$	$-\sqrt3\approx-1.732$

Concept & Theory

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

Key relations. At a constrained extremum, $\nabla f=\lambda\nabla g$ for some scalar $\lambda$ (the gradients are parallel because moving tangentially along the constraint surface produces no first-order change in $f$ unless $\nabla f$ has a normal component). Equivalently, rewriting the constraint as $x^2+y^2+(2z)^2=1$ and $f=1\cdot x+1\cdot y+(-1)\cdot(2z)$, the Cauchy–Schwarz inequality gives $|f|\le\sqrt{1^2+1^2+(-1)^2}\cdot\sqrt{x^2+y^2+(2z)^2}=\sqrt3$, with equality exactly at the Lagrange critical points.

Why this works. The ellipsoid $x^2+y^2+4z^2=1$ is closed and bounded (compact), so by the extreme value theorem the continuous function $f$ attains both 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. The Cauchy–Schwarz cross-check confirms the same bound without solving the multiplier equations, a useful sanity check whenever the objective is linear and the constraint is an ellipsoid.

Common pitfall. Reporting only one critical point (e.g. just the maximum) instead of both signs of $\lambda$; also an arithmetic slip combining $\tfrac{2}{4\lambda^2}+\tfrac{4}{16\lambda^2}$ over a common denominator, or mismatching the coefficient $4$ on $z^2$ when substituting $z=-1/(4\lambda)$ back into the constraint.

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

Question 4 of 8: Constrained Max/Min via Lagrange Multipliers (Ellipsoid Constraint)

Question 4: Constrained Max/Min via Lagrange Multipliers (Ellipsoid Constraint) (20 marks)