Question

Determine os valores de máximos, mínimos e ponto de sela: f(x,y) = 2x^3+xy²+5x²+y²

Answer 1

Encontramos esses pontos quando igualamos derivada parcial de x e derivada parcial de y a 0:

$f_x=6x^2+y^2+10x\\\\f_y=2xy+2y\\\\\\ \left \{ {{6x^2+y^2+10x=0} \atop {2xy+2y=0}} \right. \\\\\\ * 2xy+2y=0\\\\2y(x+1)=0\\\\y=0,\ x=-1\\\\\\ * 6x^2+y^2+10x=0\\\\6(-1)^2+y^2+10(-1)=0\\\\y^2=4\\\\y=\pm2\\\\6x^2+(0)^2+10x=0\\\\x=0,\ x=-\dfrac53\\\\\\S:\left\{(-1,2),(-1,-2),(0,0),\left(0,-\dfrac53\right)\right\}$

Agora que temos os pontos podemos determinar quem eh o que com o teste da segunda derivada:

$\text{max:}\\\\f_{xx}\ \textless \ 0\text{ and }(f_{xx})(f_{yy})-(f_{xy})^2\ \textgreater \ 0\\\\\\\text{min:}\\\\f_{xx}\ \textgreater \ 0\text{ and }(f_{xx})(f_{yy})-(f_{xy})^2\ \textgreater \ 0\\\\\\\text{saddle:}\\\\(f_{xx})(f_{yy})-(f_{xy})^2\ \textless \ 0\\\\\\f_{xx}=12x+10\\\\f_{yy}=2x+2\\\\f_{xy}=2y\\\\\\(-1,2):\\\\f_{xx}=12(-1)+10=-2<0\\\\f_{yy}=2(-1)+2=0\\\\f_{xy}=2(2)=4\\\\(-2)(0)-(4)^2=-16<0\text{ (saddle)}\\\\\\(-1,-2):\\\\f_{xx}=12(-1)+10=-2<0\\\\f_{yy}=2(-1)+2=0\\\\f_{xy}=2(-2)=-4\\\\(-2)(0)-(-4)^2=-16<0\text{ (saddle)}\\\\\\(0,0):\\\\f_{xx}=12(0)+10=10>0\\\\f_{yy}=2(0)+2=2\\\\f_{xy}=2(0)=0\\\\(10)(2)-(0)^2=20>0\text{ (min)}\\\\\\\left(0,-\dfrac53\right)\\\\f_{xx}=12(0)+10=10>0\\\\f_{yy}=2(0)+2=2\\\\f_{xy}=2\left(-\dfrac53\right)=-\dfrac{10}{3}\\\\(10)(2)-\left(-\dfrac{10}{3}\right)^2=\dfrac{80}{9}>0\text{ (min)}$