How to find $R$ if the maximum value of $x-y+z$ under the restriction $x^2+y^2+z^2=R^2$ is $\sqrt{27}$?

Question

Find $R$ if the maximum value of $x-y+z$ under the constraint $x^2+y^2+z^2=R^2$ is $\sqrt{27}$.

This is an extrema finding problem so we can use Lagrange multipliers here. $$ \begin{cases} 1=c2x\\ -1=c2y\\ 1=c2z \end{cases} $$ Then: $$ x=z\Leftrightarrow x^2+y^2+x^2=R^2\Leftrightarrow2x^2+y^2=R^2\\ \Leftrightarrow y=\pm\sqrt{R^2-2x^2} $$ We can plug the findings into $x-y+z$: $$ x\pm\sqrt{R^2-2x^2}+x=2x\pm\sqrt{R^2-2x^2} $$ But it doesn't seem to lead me anywhere. How to I proceed?

Answer 1

You could solve all 3 equations symmetrically as $$c=\frac{1}{2x}=-\frac{1}{2y}=\frac{1}{2z}. $$ This gives $$x=-y=z, $$ and the equation of the sphere gives $$3x^2=R^2 \implies x= \pm R/\sqrt3 .$$ The two critical points are then $$ \pm R/\sqrt{3}(1,-1,1).$$ I'll leave it to you to find which one is the maximum.

Answer 2

This is just use of Cauchy inequality: $$(x-y+z)^2\leq 3(x^2+y^2+z^2)= 3R^2$$ Maximum is reachead when $x:-y:z = 1:1:1$ thus $x=\sqrt{3}$ and $R=3$.

Answer 3

Using the substitution

$$x=R\cos u\sin v, y=-R\cos u\cos v, z=R\sin u$$ where $\displaystyle u,v\in[0,\frac\pi2]$.

(Note that $y$ is chosen to be negative so that it can give maximal $x-y+z$.)

\begin{align} x-y+z &= R(\cos u\sin v+\cos u \cos v+\sin u)\\ &=R\left(\sqrt2\cos u\sin(v+\frac\pi4)+\sin u\right)\\ &\le R(\sqrt2\cos u+\sin u )=f(u) \end{align}

Differentiate w.r.t $u$, $$f'(u)=R(\sqrt2\sin u-\cos u)$$ $$f'(u)=0\iff\tan u =\frac1{\sqrt2}\implies\cos u=\frac{\sqrt2}{\sqrt3},\sin u=\frac{1}{\sqrt3}$$

Therefore $$\sqrt{27}=R\left(\sqrt2\cdot\frac{\sqrt2}{\sqrt3}+\frac1{\sqrt3}\right)\implies R=3$$