# maximum and minimum value of $\frac{x^2+y^2}{x^2+xy+4y^2}$

If $$x,y\in\mathbb{R}$$ and $$x^2+y^2>0.$$ Then maximum and minimum value of $$\displaystyle \frac{x^2+y^2}{x^2+xy+4y^2}$$

Plan

Let $$K=\frac{x^2+y^2}{x^2+xy+4y^2}$$

$$Kx^2+Kxy+4Ky^2=x^2+y^2\Rightarrow (4K-1)y^2+Kxy+(K-1)x^2=0$$

put $$y/x=t$$ and equation is $$(4K-1)t^2+Kt+(K-1)=0$$

How do i solve it Help me plesse

• And, just graphing it, only when $K$ is equal to $1$, does anything show up. When $K$ is equal to $1$, there are no minima or maxima. Jun 13, 2019 at 0:19
• It is interesting that the function is like the reciprocal of the Rayleigh coefficient of a certain matrix, that would help. Indeed, the minimum and maximum of this function will be related with the eigenvalues of the mateix. Give me some time, I'll give it a try. Jun 13, 2019 at 0:56
• @Quote Davis Try graphing $K$ as a function of $t$. Jun 13, 2019 at 3:55

$$K=\frac{1+t^2}{1+t+4t^2}$$ To find min and max, set $$K'(t)=0$$ or $$t^2-6t-1=0$$ and solve for $$t$$. Answer $$t=3\pm \sqrt{10}$$ leading to max and min $$K=\frac{20\pm 6\sqrt{10}}{80\pm 25\sqrt{10}}$$. or $$1.088303688022443$$ and $$0.245029645310883$$

Corrected (stupid error in solving quadratic).

• How did you know to do this? Just insight or a traditional technique? Jun 13, 2019 at 2:43
• @Chase Ryan Taylor standard calculus Jun 13, 2019 at 3:02
• I meant the substitution Jun 13, 2019 at 3:41
• I don't understand your question. Note the correction. Jun 13, 2019 at 3:46
• What I mean is: For a general $f(x,y)$ to maximize, what is the formula for $K(t)$? Jun 13, 2019 at 4:17

Let $$v = (x,y)$$, $$A = \begin{bmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 4 \end{bmatrix}$$ and define the Rayleigh cuocient. $$R_A(v) = \frac{v^T A v}{v^Tv}.$$ Then, it is clear that $$\frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{R_A(v)}.$$ It is well-known (Linear Algebra result) that $$\operatorname{min}_{v \neq 0} R_A(v) = \lambda_{\operatorname{min}}$$ and $$\operatorname{max}_{v \neq 0} R_A(v) = \lambda_{\operatorname{max}}$$, where $$\lambda_{\operatorname{min}}$$ and $$\lambda_{\operatorname{max}}$$ are the smallest and largest eigenvalues of $$A$$, respectively.

In this case $$\lambda_{\operatorname{min}} = \frac{1}{2} \left(5-\sqrt{10}\right)$$ and $$\lambda_{\operatorname{max}} = \frac{1}{2} \left(5+\sqrt{10}\right)$$. Therefore, $$\min_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\max_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5+\sqrt{10}\right)} = \frac{2}{15}(5+\sqrt{10})$$ and $$\max_{x,y\neq 0} \frac{x^2+y^2}{x^2+xy+4y^2} = \frac{1}{\min_{v\neq 0} R_A(v)} = \frac{1}{\frac{1}{2} \left(5-\sqrt{10}\right)} = \frac{2}{15}(5-\sqrt{10}).$$

• +1: Nice approach. Jun 13, 2019 at 16:34

From the restriction $$x^2+y^2>0$$, we get that $$x,y$$ are not both zero.

If $$x=0$$, then $$K={\large{\frac{1}{4}}}$$.

Suppose $$x\ne 0$$.

Letting $$t={\large{\frac{y}{x}}}$$, and following your approach, we get $$(4K-1)t^2+Kt+(K-1)=0$$ which has a real solution for $$t$$ if and only $$K={\large{\frac{1}{4}}}$$ or the discriminant $$K^2-4(4K-1)(K-1)$$ is nonnegative.

Equivalently, either $$K={\large{\frac{1}{4}}}$$ or $$-15K^2+20K-4\ge 0$$ With the restriction $$K\ne{\large{\frac{1}{4}}}$$, the quadratic inequality solves as $$\frac{10-2\sqrt{10}}{15}\le K\le \frac{10+2\sqrt{10}}{15},\;\;K\ne{\large{\frac{1}{4}}}$$ Noting that $$\frac{10-2\sqrt{10}}{15}<\frac{1}{4}<\frac{10+2\sqrt{10}}{15}$$ it follows that

• The minimum value of $$K$$ is $${\large{\frac{10-2\sqrt{10}}{15}}}\approx .2450296455$$.$$\\[8pt]$$
• The maximum value of $$K$$ is $${\large{\frac{10+2\sqrt{10}}{15}}}\approx 1.088303688$$.