# Maximum and minimum of a fractional function

Let $x, y \in \mathbb{R}$, $a, b, c$ are three real parameters with $c\neq 0$. Find the maximum and minimum of $\dfrac{ax+by+c}{\sqrt{x^2+y^2+1}}$

This is quite complicated if I calculate the derivative. Is there any other ways? Please help me.

Thanks.

I know that some people has voted my question down, I know how to use Cauchy-Schwarz inequality, but this only gives me the maximum, not the minimum. I'm not good at this kind of math, so, instead of voting down, please explain for me.

• Define $u=(x,y,1)$ and $v=(a,b,c)$. Then you're just looking at $u\cdot v/|u|$... – Tom-Tom Jan 30 '15 at 15:16
• Cauchy-Schwarz gives you the minimum as well – r9m Jan 30 '15 at 15:41

Your function, call it $F$, is $c$ times the dot product of $(a/c,b/c,1)$ and the unit vector $\vec{n}$ in the direction of $(x,y,1)$. If $\theta$ is the angle between $(a/c,b/c,1)$ and $\vec{n}$, then $(a/c,b/c,1) \cdot \vec{n}$ is minimized when $\cos\theta = -1$, in which case the minimum value of $F$ is $c(-\|(a/c,b/c,1)\|) = -\sqrt{a^2 + b^2 + c^2}$, if $c > 0$. If $c < 0$, the minimum value of $F$ is $c$ times the maximum of $(a/c,b/c,1) \cdot \vec{n}$. Since $(a/c,b/c,1)\cdot \vec{n}$ is maximized when $\cos\theta = 1$, the minimum of $F$ is $c\|(a/c,b/c,1)\| = -\sqrt{a^2 + b^2 + c^2}$.
• It's actually not so easy. Since the last component of $\vec{n}$ is positive, the Cauchy-Schwarz inequality gives you only one, the maximum if $c > 0$, and the minimum if $c < 0$. – Daniel Fischer Jan 30 '15 at 15:57