Extreme values of function with two variables $z=\frac{ax+by+c}{\sqrt{x^2+y^2+1}}$

Question

I've been struggling with following problem:

Determine extreme values of function $$ z = \frac{ax+by+c}{\sqrt{x^2+y^2+1}}, $$ where $a^2+b^2+c^2 \neq0$.

Here is my approach:

As we know differential function can have extreme value only in stationary points, i.e. in point where derivative is zero. So let's calculate partial derivatives of given function. We have: $$ \frac{\partial z}{\partial x} = \frac{a\sqrt{x^2+y^2+1} - \frac{ax+by+c}{2\sqrt{x^2+y^2+1}}2x}{x^2+y^2+1} = \frac{ay^2+a-bxy-cx}{(x^2+y^2+1)^{\frac{3}{2}}} $$ and $$ \frac{\partial z}{\partial y} = \frac{bx^2+b-axy-cy}{(x^2+y^2+1)^{\frac{3}{2}}} $$

So now we need to solve system of equations \begin{align} &\begin{cases} \dfrac{\partial z}{\partial x} = 0 \\[3pt] \dfrac{\partial z}{\partial y} = 0 \end{cases} \\[5pt] \implies &\begin{cases} \hfil \dfrac{ay^2+a-bxy-cx}{(x^2+y^2+1)^{\frac{3}{2}}} = 0 \\[3pt] \hfil \dfrac{bx^2+b-axy-cy}{(x^2+y^2+1)^{\frac{3}{2}}} = 0 \end{cases} \\[5pt] \implies &\begin{cases} \hfil ay^2+a-bxy-cx = 0 \\[3pt] \hfil bx^2+b-axy-cy = 0 \end{cases} \end{align}

Now I've been trying to solve this system of equations for long time but I just cannot do it :( It feels like there should be some clever way to solve this system, but I just don't see it.

Answer 1

Hint:

By the Cauchy-Schwarz Inequality, we have that: $$|ax+by+c| \leq |a| \cdot|x| + |b| \cdot |y| + |c| \cdot 1 \leq \sqrt{a^2+b^2+c^2} \sqrt{x^2+y^2+1}$$

Try doing some algebraic gymnastics with this :) This may still require some work but it should be easier to handle.

Answer 2

$$ay^2+a-bxy-cx = 0 \tag 1$$ $$bx^2+b-axy-cy = 0 \tag 2$$

From $(1)$,you have $$x=\frac {a(y^2+1)}{by+c}$$ Plug $x$ in $(2)$ and simplify to obtain $$\frac{(b-c y) \left(a^2 \left(y^2+1\right)+(b y+c)^2\right)}{(b y+c)^2}=0$$ So, there is one solution $y=\frac bc$ to which corresponds $x=\frac ac$.

What remains is to look at $$a^2 \left(y^2+1\right)+(b y+c)^2= \left(a^2+b^2\right)y^2+2 b c y+\left(a^2+c^2\right)=0$$ which has no real root since $$\Delta=-4a^2(a^2+b^2+c^2) ~~< ~0$$