Maximum and minimum value of $\frac{8x - 3y}{\sqrt{4x^2+y^2}}$?

Question

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.

Answer 1

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.

Answer 2

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 $-5$.