Finding maxima of multivariable functions in bounded regions Find the maximum and minimum values of $xz+yz$ over the ball $x^{2}+y^{2}+z^{2} \leq 1$

I frequently have trouble solving these kind of exercises that ask you to find extrema on a bound region. How can I go over them? My textbook offers a solution, but it's extremely complicated involving a polar coordinate change. The partial derivatives are the following:
$f_{x} = z$
$f_{y} = z$
$f_{z} = x+y$
After this, how can I obtain the maxima in the region that I'm given? (in this case the sphere). Is there to say, a more "general" method to solve these kind of exercises? Thank you very much for your time.
 A: Lagrange multipliers are really useful in order to find constrained maxima/minima.
In particular, the stationary points of a differentiable function $f:\mathbb{R}^n\to\mathbb{R}$ over the unit ball are located in points that satisfy:
$$ \forall i\in[1,n],\quad \frac{\partial f}{\partial x_i}=\lambda x_i.$$
Since your function is a differentiable and homogeneous one, you can clearly assume that the stationary points are located on the unit ball. Moreover, since $f_x=f_y$, in virtue of Lagrange's Theorem all the stationary points lie on the hyperplane $x=y$: this decreases the "dimension" of the problem - you simply have to find the stationary points of $2xz$ over the ellipse $2x^2+z^2=1$. 
Just like the original problem, this can be done through the AM-GM or Cauchy-Schwarz inequality, for example by setting $x=\frac{w}{\sqrt{2}}$, from which you have to find maxima/minima of $\sqrt{2} wz$ over $w^2+z^2=1$: they clearly are $\pm\frac{1}{\sqrt{2}}$. 
Otherwise, you can take a parametrization of the unit ball in terms of Euler angles:
$$ x = \sin\phi \sin\theta,\quad y=\sin\phi\cos\theta,\quad z=\cos\phi $$
and find (probably faster) maxima and minima of:
$$ z(x+y) = \cos\phi\sin\phi(\sin\theta+\cos\theta) = \frac{1}{\sqrt{2}}\sin(2\phi)\sin(\theta+\pi/4).$$
A: since
$$xz+yz=z(x+y)\le|z(x+y)|\le\sqrt{2}\cdot \sqrt{z^2(x^2+y^2)}\le\sqrt{2}\sqrt{\dfrac{(z^2+x^2+y^2)^2}{4}}\le\dfrac{\sqrt{2}}{2}$$
if and only if $$|x|=|y|,|z|^2=x^2+y^2=\dfrac{1}{2}$$
