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I have to study the type of critical points of the function $$ f(x,y)=(2x^2+y^2-1)(x^2+y^2-1)+1 $$ and find minimum and maximum on the generic circle centered in $ (0,0) $ and radius $ r>1 $.

I did the partial derivates and tried to solve this system $$ \left\{\begin{matrix} f_{x}=8x^3-6x+6xy^2=0\\ f_{y}=4y^3-4y+6xy^2=0 \end{matrix}\right. $$ But I don't know how to solve this.

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  • $\begingroup$ $8x^3-6x+6xy^2=x(8x^2-6+6y^2)=0 \implies x=0 ,8x^2-6+6y^2=0$ $\endgroup$ – chenbai Feb 6 '16 at 9:47
  • $\begingroup$ With $ x=0 $ I found the critical points $(0,0),(0,-1),(0,1)$ but how can I use the other condition? $\endgroup$ – Aoeilda Feb 6 '16 at 10:17
  • $\begingroup$ you can check the points for $x^2+y^2>1$, if not, they are not the points you want. also you need to know if they are local min or local max. $\endgroup$ – chenbai Feb 6 '16 at 12:56
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Hint

You are asked to find the maximum and minimum of $f(x,y)=(2x^2+y^2-1)(x^2+y^2-1)+1$ subject to the constraint $x^2+y^2=r^2.$ This can be done using Lagrange multipliers. See https://en.wikipedia.org/wiki/Lagrange_multiplier

You have to consider $L(x,y,\lambda)=(2x^2+y^2-1)(x^2+y^2-1)+1+\lambda (x^2+y^2-r^2).$ Now, the constrained extrema of $f$ are critical points of $L.$ So, one has that solve

$$\begin{cases}L_x=0 \\ L_y=0 \\ L_{\lambda}=0\end{cases}$$ The system can be a bit complicated. Now, note that because of the presence of $x^2+y^2-1$ which is similar to $x^2+y^2-r^2$ we can write $f(x,y)$ as

$$(2x^2+y^2-1)(x^2+y^2-1)+(2x^2+y^2-1)(x^2+y^2-r^2)-(2x^2+y^2-1)(x^2+y^2-r^2)+1.$$ That is, using the constraint,

$$f(x,y)=(2x^2+y^2-1)(r^2-1)+1,$$ which much easier. So, apply Lagrange multiplies to get maximum/minimum values.

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In your special case the minima and maxima can be computed without using Lagrange multipliers or partial derivatives.

For all points $(x, y)$ on the circle with center $(0, 0)$ and radius $r$ you have $x^2 + y^2 = r^2$ and therefore $$ f(x, y) = (x^2 + r^2 - 1)(r^2 - 1) + 1 \, . $$ Every term apart is constant apart from $x \in [-r, r]$, and the factor$r^2 -1$ is positive. Therefore $f$ is minimal for $x = 0$ (which implies $y = \pm r$) and maximal for $x = \pm r$ (which implies $y = 0$).

The same result can be obtained by using polar coordinates: $$ g(\theta) = f(r \cos \theta, r \sin \theta) = (r^2 \cos^2 \theta + r^2 - 1)(r^2 -1)+1 $$ is minimal/maximal exactly where $\cos^2 \theta $ is.

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