Extrema of $f(x,y)=(1-x^2-y^2)\cdot xy$ subject to $x^2+y^2\leq 1$ I want to determine the extrema of $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ subject to $x^2+y^2\leq 1$.
We use Lagrange Multipliers to check the critical points on the circle $x^2+y^2=1$.
To check the critical  points inside the circle, $x^2+y^2<1$, we calculate the critical points of $f(x,y)$.
Let's start with the Lagrange-Multipliers.
We have $f(x,y)=(1-x^2-y^2)\cdot xy=xy-x^3y-xy^3$ und $g(x,y)=x^2+y^2-1=0$.
We consider
\begin{equation*}L(x,y,\lambda )=f(x,y)+\lambda g(x,y)=xy-x^3y-xy^3+\lambda \left (x^2+y^2-1\right )\end{equation*}
We calculate the partial derivatives of $L$.
\begin{align*}&\frac{\partial{L}}{\partial{x}}=y-3x^2y-y^3+2\lambda x \\ &\frac{\partial{L}}{\partial{y}}=x-x^3-3xy^2+2\lambda y \\ &\frac{\partial{L}}{\partial{\lambda }}= x^2+y^2-1 \end{align*}
To calculate the extrema we set each equation equal to zero and solve the system:
\begin{align*}&y-3x^2y-y^3+2\lambda x=0 \ \ \ \ \ \ \ \ (1) \\ & x-x^3-3xy^2+2\lambda y  =0 \ \ \ \ \ \ \ \ (2) \\ &  x^2+y^2-1=0 \ \ \ \ \ \ \ \ (3)\end{align*}
We get the critical points \begin{equation*}P_1\left (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right ) \ \ , \ \ \ P_2\left (-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right )\ \ , \ \ \ P_3\left (-1, 0\right )\ \ , \ \ \ P_4\left (1, 0\right ) \ \ , \ \ \ P_5\left (0, -1\right ) \ \ , \ \ \ \ P_6\left (0, 1\right )\end{equation*}
Now we check inside the circle.
We set the partial derivatives of $f$ equal to zero and solve the system:
\begin{align*}&f_x=0 \Rightarrow y-3x^2y-y^3=0 \ \ \ \ \ \ \ \ (1)\\ &f_y=0 \Rightarrow x-x^3-3xy^2=0\ \ \ \ \ \ \ \ (2)\end{align*}
Then we get the critical points \begin{align*}&Q_1\left (0, 0\right ) \ \ , \ \ \ Q_2\left (\frac{1}{2}, \frac{1}{2}\right )\ \ , \ \ \ Q_3\left (-\frac{1}{2}, -\frac{1}{2}\right )\ \ , \ \ \ Q_4\left (0, -1\right )\ \ , \ \ \ Q_5\left (0, 1\right )\ \ , \ \ \ Q_6\left (-2, -1\right ) \ \ , \ \ \ Q_7\left (0, -1\right )\ \ , \ \\ & \ Q_8\left (-\frac{3}{2}, -\frac{1}{2}\right ) \ \ , \ \ \ Q_9\left (\frac{1}{2}, -\frac{1}{2}\right )\  \ , \ \ \ Q_{10}\left (2, 1\right )\ \ , \ \ \ Q_{11}\left (\frac{3}{2}, \frac{1}{2}\right ) \ \ , \ \ \ Q_{12}\left (-\frac{1}{2}, \frac{1}{2}\right )\end{align*}
Then we have to find the maximal and minimum value of $f$ at all these critical points.
Is everything correct? I am not really sure about the part inside the circle, if all critical points are correct.

Let $x=r\cos t$ and $y=r\cos t$, where $0\leq r\leq1$ and $t\in[0,2\pi)$.
Thus, by AM-GM $$(1-x^2-y^2)xy=(1-r^2)r^2\sin t\cos t=\frac{1}{2}(1-r^2)r^2\sin2t\leq$$
$$\leq\frac{1}{2}(1-r^2)r^2\leq\frac{1}{2}\left(\frac{1-r^2+r^2}{2}\right)^2=\frac{1}{8}.$$
The equality occurs for $1-r^2=r^2,$ which gives $r=\frac{1}{\sqrt2}$ and $\sin2t=1,$ which says that we got a maximal value.
By the similar way we obtain that $-\frac{1}{8}$ is a minimal value.
Another way.
By AM-GM twice we obtain:
$$(1-x^2-y^2)xy\leq(1-x^2-y^2)\cdot\frac{1}{2}(x^2+y^2)\leq\frac{1}{2}\left(\frac{1-x^2-y^2+x^2+y^2}{2}\right)^2=\frac{1}{8}.$$
The equality occurs for $x=y$ and $1-x^2-y^2=x^2+y^2,$ which says that we got a maximal value.
Also, $$(1-x^2-y^2)xy\geq(1-x^2-y^2)\cdot\left(-\frac{1}{2}(x^2+y^2)\right)\geq$$
$$\geq-\frac{1}{2}\left(\frac{1-x^2-y^2+x^2+y^2}{2}\right)^2=-\frac{1}{8}.$$
The equality occurs for $x=-y$ and $1-x^2-y^2=x^2+y^2,$ which says that we got a minimal value.
