Exercise 6.1.7

Take derivative of $R$ w.r.t. $x,y$, we have

$\frac{\mathit{\partial R}}{\mathit{\partial x}}=\frac{-2x{y}^2}{{(x^2+y^2)}^2}$, $\frac{\mathit{\partial R}}{\mathit{\partial y}}=\frac{2y{x}^2}{{(x^2+y^2)}^2}$

The second derivatives are:

$H_{11}=\frac{{\partial }^{2}R}{\partial x^2}=\frac{-2y^2{(x^2+y^2)}^2-2(x^2+y^2)2x(-2x{y}^2)}{{(x^2+y^2)}^4}=\frac{(x^2+y^2)y^2\left(-2(x^2+y^2)+8x^2\right)}{{(x^2+y^2)}^4}=\frac{2y^2\left(3x^2-y^2\right)}{{(x^2+y^2)}^3}$

$H_{22}=\frac{{\partial }^{2}R}{\partial y^2}=\frac{2x^2{(x^2+y^2)}^2-2(x^2+y^2)2y(2y{x}^2)}{{(x^2+y^2)}^4}=\frac{(x^2+y^2)x^2\left(2(x^2+y^2)-8y^2\right)}{{(x^2+y^2)}^4}=\frac{2y^2\left(x^2-3y^2\right)}{{(x^2+y^2)}^3}$

$H_{12}=H_{21}=\frac{{\partial }^{2}R}{\mathit{\partial x\partial y}}=\frac{4\mathit{xy}{(x^2+y^2)}^2-2(x^2+y^2)2x(2y{x}^2)}{{(x^2+y^2)}^4}=\frac{(x^2+y^2)\mathit{xy}\left(4(x^2+y^2)-8x^2\right)}{{(x^2+y^2)}^4}=\frac{4\mathit{xy}\left(y^2-x^2\right)}{{(x^2+y^2)}^3}$

We see that $R=\frac{x^2+2y^2}{x^2+y^2}=1+\frac{y^2}{x^2+y^2}$, and it’s clear that the minimum is $R=1$ when $y=0$, and the maximum value is $R=2$ when $x=0$.

Take $x=0$ into above derivatives, we have $H=\left[\begin{array}{cc}\hfill \frac{-2}{y^2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{-6}{y^2}\hfill \end{array}\right]$, which is negative definite.