I'm quite stuck in solving this exercise:
$$f(x, y, z) = ax + by + cz \qquad \qquad on \qquad \qquad x^2+y^2+z^2 \leq R^2$$
It's basically a normal problem, we study internal points, and then we study what happens on the boundary, except for the fact that there are parameters $a, b, c$ and $R$ that are general (with logic).
I started with internal points but there is no sense in this for I would get $a = b = c = 0$.
On the bounary, for example: $z = \sqrt{R^2-y^2-z^2}$ and hence
$$f(y, z) = a\sqrt{R^2-y^2-z^2} + by + cz$$
From here I would go on with studying $\nabla f = 0$ whith turns to be quite a mess: $$\begin{cases} -y + ab\sqrt{R^2-y^2-z^2} = 0 \\\\ -z + cb\sqrt{R^2 - y^2-z^2} = 0 \end{cases} $$
I thought to write $y^2$ in terms of $a, b, c, R, z$ which is possible, but then while finding $z$ I get
$$z = \pm \frac{cbR}{\sqrt{1 + a^2b^2 + c^2b^2}}$$
Well, from here I could find out $y$ by going back, but then I would have two points depending on $a, b, c$.
Is there a better way to solve this problem?? The request is always to find max / min of $f$ over that ball.
Hint.
The ball is a closed compact and $f$ is linear so the minimum/maximum are located on the boundary. Using Lagrange multipliers the lagrangian reads
$$ L(x,y,z,\lambda) = ax + by +c z+ \lambda(x^2+y^2+z^2-R^2) $$
the stationary points are the solutions for
$$ \nabla L = 0 = \cases{a + 2\lambda x\\ b+2\lambda y\\ c +2\lambda z\\ x^ 2+y^2+z^2 -R^2} $$