Maxima and Minima over $\mathbb{R^3}$

Question

Let $a, b, c$ and $d$ be real numbers. Find, with full justifications, the maximum and minimum values of the function $$ f(x, y, z)=\frac{a x+b y+c z+d}{1+x^{2}+y^{2}+z^{2}} $$ over $\mathbb{R}^{3}$

For maximum value denominator has to minimum so ${1+x^{2}+y^{2}+z^{2}}$ minimum value is $1$ and hence maximum of function is $d$. Is this correct?

Answer 1

Yes, the minimum of $1+x^2+y^2+z^2$ is $1$ and it is attained at $(0,0,0)$, but why should the numerator have a maximum at that point? It hasn't.

Usng the fact that$$\lim_{\lVert(x,y,z)\rVert\to\infty}f(x,y,z)=0,$$you can solve the problem by determining the points at which $\nabla f(x,y,z)=(0,0,0)$. By doing this, you can prove that the maximum and the minimum of $f$ are $\frac12\left(d\pm\sqrt{a^2+b^2+c^2+d^2}\right)$.

Answer 2

Here is a way how to reduce your problem to a $1$-dimensional one which then can be easily solved:

• Let's denote $u = \begin{pmatrix} a \\ b \\ c \end{pmatrix}$ and $v = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$

I will consider only the case $u\neq \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$.

Using $u\cdot v$ to denote the dot-product, your function looks now as follows:

$$f(v) = \frac{u\cdot v + d}{1+||v||^2}$$

Applying Cauchy-Schwarz inequality to the dot product gives

$$\frac{- ||u|| ||v|| +d}{1+||v||^2} \leq f(v) \leq \frac{||u|| ||v|| + d}{1+||v||^2}$$

Here, equality is only reached for $t \neq 0$ and $v = tu$. So,

$$\frac{d- |t|||u||^2 }{1+t^2||u||^2} \leq f(v) \leq \frac{d +|t|||u||^2}{1+t^2||u||^2}$$

Now, just study the function $g(t) =\frac{d +t||u||^2}{1+t^2||u||^2}$ wrt. to global maxima and minima. (I leave this part up to you.)

You will find that $g$ has a global maximum $\boxed{\frac{1}{2}\left(d+ \sqrt{d^2+||u||^2}\right)}$ at $t_{max} = \frac{1}{||u||^2}\left(-d+ \sqrt{d^2+||u||^2}\right)$ and a global minimum $\boxed{\frac{1}{2}\left(d- \sqrt{d^2+||u||^2}\right)}$ at $t_{min} = \frac{1}{||u||^2}\left(-d- \sqrt{d^2+||u||^2}\right)$.