Question 2 of 8: Extrema of a Linear Function on a Sphere

Question 2: Extrema of a Linear Function on a Sphere (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 $f(x,y,z)=x+y-z$; constraint $g=x^2+y^2+z^2-1=0$ (the unit sphere, compact).

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

Approach. The unit sphere is closed and bounded, so $f$ attains both extrema on it. Apply Lagrange multipliers: $\nabla f=\lambda\nabla g$ together with the constraint gives exactly two critical points.

1. Lagrange conditions. $\nabla f=(1,1,-1)$, $\nabla g=(2x,2y,2z)$, so $$1=2\lambda x,\quad 1=2\lambda y,\quad -1=2\lambda z\ \Rightarrow\ x=y=\frac{1}{2\lambda},\quad z=-\frac{1}{2\lambda}.$$
2. Substitute into the constraint. $$x^2+y^2+z^2=\frac{3}{4\lambda^2}=1\ \Rightarrow\ \lambda^2=\frac34\ \Rightarrow\ \lambda=\pm\frac{\sqrt3}{2}.$$
3. Evaluate $f$ at both critical points. $f=x+y-z=\dfrac{3}{2\lambda}$. For $\lambda=+\tfrac{\sqrt3}{2}$: $f=\sqrt3$ (maximum, at $(x,y,z)=(\tfrac{1}{\sqrt3},\tfrac{1}{\sqrt3},-\tfrac{1}{\sqrt3})$). For $\lambda=-\tfrac{\sqrt3}{2}$: $f=-\sqrt3$ (minimum, at the antipodal point).

$$f_{\max}=\boxed{\sqrt3},\qquad f_{\min}=\boxed{-\sqrt3}$$