The following sentences are from Michael Spivak's "Calculus on Manifolds".
2-6 Theorem. Let $A\subset\mathbb{R}^n$. If the maximum (or minimum) of $f:A\to\mathbb{R}$ occurs at a point $a$ in the interior of $A$ and $D_if(a)$ exists, then $D_if(a)=0.$
If Theorem 2-6 is used to find the maximum or minimum of $f$ on $A$, the values of $f$ at boundary points must be examined separately - a formidable task, since the boundary of $A$ may be all of $A$! Problem 2-27 indicates one way of doing this,
2-27. Define $g,h:\{x\in\mathbb{R}^2: |x|\leq 1\}\to\mathbb{R}^3$ by $$g(x,y)=(x,y,\sqrt{1-x^2-y^2}),\\ h(x,y)=(x,y,-\sqrt{1-x^2-y^2}).$$
Show that the maximum of $f$ on $\{x\in\mathbb{R}^3: |x|=1\}$ is either the maximum of $f\circ g$ or the maximum of $f\circ h$ on $\{x\in\mathbb{R}^2: |x|\leq 1\}.$
How to use the result of Problem 2-27 to find the maximum of $f$?
My idea:
Suppose that $f$ is differentiable on $\{x\in\mathbb{R}^3: |x|\leq 1\}$.
Suppose that we want to find the maximum of $f$ on $\{x\in\mathbb{R}^3: |x|\leq 1\}$.
Find the set $S$ of all the points $a$ from $\{x\in\mathbb{R}^3: |x|< 1\}$ such that $D_if(a)=0.$
Assume that $S$ is finite.
Let $m_1:=\max\{f(a):a\in S\}$.
By Problem 2-27, the maximum of $f$ on $\{x\in\mathbb{R}^3: |x|=1\}$ is either the maximum of $f\circ g$ or the maximum of $f\circ h$ on $\{x\in\mathbb{R}^2: |x|\leq 1\}.$
We reduced the original three-dimensional problem ($\max f$ on $\{x\in\mathbb{R}^3: |x|\leq 1\}$) to two two-dimensional problems ($\max f\circ g$ on $\{x\in\mathbb{R}^2: |x|\leq 1\}$ and $\max f\circ h$ on $\{x\in\mathbb{R}^2: |x|\leq 1\}$).
Similarly, we can reduce these two two-dimensional problems to one-dimensional problems on $[-1,1]$.
By this reduction, we can find $m_2:=\max\{f(a):a\in\{x\in\mathbb{R}^3:|x|=1\}\}$.
The answer of this problem is $\max\{m_1,m_2\}$.
This is how to use the result of Problem 2-27 to find the maximum of $f$.
Is my idea ok?