Extrema of a Linear Function on a Sphere via Lagrange Multipliers: 04-BS-1, May 2017, Q3

Question 3: Extrema of a Linear Function on a Sphere 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. A linear objective $f(x,y,z)=x+y-z$ on the compact constraint surface $g=x^2+y^2+z^2-1=0$ (the unit sphere, so global extrema are guaranteed).

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

Approach. Solve $\nabla f=\lambda\nabla g$ with the constraint. Since $f$ is linear, the result reproduces the Cauchy–Schwarz bound $|f|\le\|\nabla f\|$ — a useful independent check.

Set up the Lagrange conditions. $\nabla f=(1,1,-1)$, $\nabla g=(2x,2y,2z)$, so $$1=2\lambda x,\qquad 1=2\lambda y,\qquad -1=2\lambda z\ \Rightarrow\ x=y=\dfrac{1}{2\lambda},\quad z=-\dfrac{1}{2\lambda}.$$
Apply the constraint. $x^2+y^2+z^2=3\left(\dfrac{1}{2\lambda}\right)^2=1\ \Rightarrow\ \lambda^2=\dfrac34\ \Rightarrow\ \lambda=\pm\dfrac{\sqrt3}{2}$.
Evaluate $f$ at each critical point. $f=x+y-z=\dfrac{3}{2\lambda}$. For $\lambda=\tfrac{\sqrt3}{2}$: $f=\dfrac{3}{\sqrt3}=\sqrt3$, at $\left(\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3}\right)$. For $\lambda=-\tfrac{\sqrt3}{2}$: $f=-\sqrt3$, at $\left(-\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3}\right)$.
Compare (the sphere is compact, so these are the global extrema). $$f_{\max}=\boxed{\sqrt3}\approx1.732,\qquad f_{\min}=\boxed{-\sqrt3}\approx-1.732$$ (Cross-check via Cauchy–Schwarz: $|f|=|(1,1,-1)\cdot(x,y,z)|\le\|(1,1,-1)\|\cdot\|(x,y,z)\|=\sqrt3\cdot1=\sqrt3$, with equality exactly at these two points — agrees.)

Quantity	Result
Critical point (max)	$\left(\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3}\right)$
$f_{\max}$	$\sqrt3\approx1.732$
Critical point (min)	$\left(-\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3}\right)$
$f_{\min}$	$-\sqrt3\approx-1.732$

Concept & Theory

Topic. Constrained extrema of a linear objective on a sphere via Lagrange multipliers.

Key relations. At an extremum of $f$ subject to $g=0$, $\nabla f=\lambda\nabla g$. For a linear objective $f=\mathbf a\cdot\mathbf x$ on the sphere $\|\mathbf x\|=r$, the extrema are exactly $\pm r\|\mathbf a\|$ by the Cauchy–Schwarz inequality — the Lagrange algebra and the inequality bound are two routes to the same answer.

Why this works. Because $\nabla f=(1,1,-1)$ is constant, all three gradient-matching equations give $x,y,z$ as the same multiple of $1/\lambda$ (up to sign), collapsing what looks like a 3-variable system to a single equation in $\lambda$ once substituted into the constraint.

Common pitfall. Forgetting the sphere's compactness is what guarantees these two critical points really are the global max/min (not just local); sign slips when substituting $z=-1/(2\lambda)$ into $f=x+y-z$.

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

Question 3 of 8: Extrema of a Linear Function on a Sphere via Lagrange Multipliers

Question 3: Extrema of a Linear Function on a Sphere via Lagrange Multipliers 20 marks