Réponse :
Bonjour,
[tex]f(x,y)=x^2+3x^2y-xy[/tex]
1) recherche des points critiques:
[tex]\left\{\begin{array}{ccc}\dfrac{\partial{f}}{\partial{x} } &=&2x+6xy-y=0\\\\\dfrac{\partial{f}}{\partial{y} } &=&3x^2-x=0\\\end{array}\right.\\\\\\\left\{\begin{array}{ccc}x&=&0\\\\y&=&0\\\\\end{array}\right.\ ou\ \left\{\begin{array}{ccc}x&=&\dfrac{1}{3}\\\\y&=&\dfrac{-2}{3} \\\\\end{array}\right.\\[/tex]
2)
[tex]\left\{\begin{array}{ccccc}\dfrac{\partial{f^2}}{\partial{x^2} } &=&2x+6y&=&r\\\\\dfrac{\partial{f^2}}{\partial{y^2} } &=&0&=&t\\\\\dfrac{\partial{f^2}}{\partial{x} \partial{y} } &=&6x-1&=&s\\\end{array}\right.\\\\\rho=rt-s^2=-(6x-1)^2 \\[/tex]
[tex]pour\ (x,y)=(0,0),\ \rho =0:\ on\ ne\ peut\ rien\ conclure\\pour\ (x,y)=(\dfrac{1}{3},\dfrac{-2}{3} ),\ \rho=-\dfrac{8}{9}\ pas\ d'extremum \:\ un\ point\ selle.[/tex]
