Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$? For real numbers $\,x, y\neq 0\,$ consider
$$\frac{8x - 3y}{\sqrt{4x^2+y^2}}\,.$$
How to find the maximum and minimum value?
I've already got the maximum by using the Cauchy–Schwarz inequality
$$\big[(2x)^2 + y^2\big]\big[4^2 + (-3)^2\big] \geq (8x - 3y)^2\\[3ex]
25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\\[4ex]
5 \geq \frac{8x - 3y}{\sqrt{(2x)^2 + y^2}}$$
But I cannot get the minimum value.
 A: You are almost there: Note that by Cauchy-Schwarz you  did
$$25 \geq \frac{(8x - 3y)^2}{(2x)^2 + y^2}\,.$$
Thus
$$\left|\frac{8x-3y}{\sqrt{4x^2+y^2}}\right|\le 5\tag {*}\\[5ex]
\implies -5\le \frac{8x-3y}{\sqrt{4x^2+y^2}}\le 5$$
The important part was in step (*). The modulus was crucial as $8x-3y$ could be negative too.
A: Let $r > 0$ and $0 \le \theta < 2\pi$, and $x = r \cos \theta$, $y = 2r \sin \theta$, so that $$4x^2 + y^2 = 4r^2,$$ hence $$\begin{align}f(x,y) &= \frac{8x-3y}{\sqrt{4x^2 + y^2}} \\ &= \frac{8r \cos \theta - 6r \sin \theta}{2r} \\ &= 4 \cos \theta - 3 \sin \theta \\
&= 5 \left(\frac{4}{5} \cos \theta - \frac{3}{5} \sin \theta \right) \\
&= 5 \left( \sin \psi \cos \theta - \cos \psi \sin \theta \right) \quad \quad \psi = \arcsin \frac{4}{5} \\
&= 5 \sin(\theta - \psi).
\end{align}$$  Now we can see this function depends only on the choice of angle $\theta$.  This is obviously maximized when $\theta = \psi + \pi/2$ and the maximum attained is $5$.
This also happens to get us the minimum for free, which occurs when $\theta = \psi + \frac{3\pi}{2},$ and the minimum is $-5$.
