Given $x^2+y^2>0$ where $x,y \in R$

Question

Find the maximum and minimum value of the following expression E

$$E=\dfrac{x^2+y^2}{x^2+xy+4y^2}$$

Answer 1

Your question can be restated as:

There exists $x,y \in \mathbb{R}$ such that $x^2+y^2>0$, i.e. $x\neq 0$ or $y\neq 0$ and $E=\dfrac{x^{2}+y^{2}}{x^{2}+x y+4 y^{2}}$, find the maximum and minimum of $E$.

Which is equivalent with:

There exists $x,y \in \mathbb{R}$ such that $x\neq 0$ or $y\neq 0$ and $(E-1)x^2+Exy+(4E-1)y^2=0$, find the maximum and minimum of $E$.

Suppose $y\neq 0$ WLOG and set $t=\dfrac{x}{y}$, we have $(E-1)t^2+Et+(4E-1)=0$ has a solution, which means the discriminant $\Delta= E^2-4(E-1)(4E-1)=-15E^2+20E-4\geqslant 0$ therefore $\dfrac{2-2\sqrt{2/5}}{3}\leqslant E \leqslant \dfrac{2+2\sqrt{2/5}}{3}$.

Answer 2

Optimizing $E$ is the same as to optimize $$ F=\frac{1}{E}=\frac{x^2+xy+4y^2}{x^2+y^2}=1+\frac{xy+3y^2}{x^2+y^2}. $$ Introduce the polar coordinates $x=r\cos\phi$, $y=r\sin\phi$. Then \begin{align} F&=1+\sin\phi\cos\phi+3\sin^2\phi=1+\frac{1}{2}(\sin2\phi+3(1-\cos2\phi))=\\ &=\frac52+\frac12(\sin2\phi-3\cos2\phi)=\frac52+\frac{\sqrt{10}}{2}\sin(2\phi+\delta) \end{align} where in the last equality the standard rewriting is done.

Now it is clear that $F$ changes between $\frac{5\pm\sqrt{10}}{2}$. For the original $E$ it corresponds to $$ \frac{2}{5\pm\sqrt{10}}=\frac{2(5\mp\sqrt{10})}{15}. $$

Answer 3

With $$y=ux$$ we have $$\dfrac{x^{2}+y^{2}}{x^{2}+x y+4 y^{2}}= \frac {1+u^2}{1+u+4u^2} =f(u)$$

$$f'(u)=0 \implies u=3\pm \sqrt {10}$$

The maximum value is $$f(3-\sqrt {10})=1.0883037..$$ and the minimum value is $$f(3+\sqrt {10})=0.245029645...$$

Answer 4

A sketch of a possible solution:

you want to find the extrema of $E(x,y)=\dfrac{x^2+y^2}{x^2+xy+4y^2}$ on $\mathbb R^2\setminus \{(0,0)\}.$

Switching to polar coordinates gives $E(r,\theta)=\frac{1}{\cos^2\theta+\sin \theta \cos \theta+4\sin^2\theta}$ so $E$ is independent of $r$ and bounded on all of $\mathbb R^2.$ So if it has extrema, they will occur along radial lines.

Set $f(\theta)=\cos^2\theta+\sin \theta \cos \theta+4\sin^2\theta$ and simplify:

$f(\theta)=\cos^2\theta+\sin \theta \cos \theta+4\sin^2\theta=1+\frac{1}{2}\sin2\theta+\frac{3}{2}(1-\cos2\theta)=$

$\frac{5}{2}+\frac{1}{2}\left(\sin2\theta-3\cos2\theta\right).$

This expression is never zero, (but we already proved this when we noted that $E$ is bounded). Now, apply the usual calculus I analysis of $f$: on $[0,2\pi],\ f$ has exactly two maxima and two minima, and relate the result to $E$ to conclude.