8/15/2019 comandos tarea 2

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\begin{document}1.a\\f(x,y)=x^2-y^2+xy\\

$\bigtriangledown f = (2x+y-2y+x)$\\Ahora $\bigtriangledown f (x,y)=0$ solo si (x,y)=(0,0) de modo que el punto critico de f es 0,0.\\

$\frac{\delta^2 f}{\delta x^2}=2+1$\\$\frac{\delta^2 f}{\delta y^2}=1-2$\\$\frac{\delta^2 f}{\delta x \delta y}=0$\\

$D=(3)(-1)= -3>0 $\\

\\\\\textbf{1.g}\\f(x,y)=(x+y)(xy+1)=x^2+x+xy^2+y\\$\bigtriangledown f (2xy+1+y^2+x^2+2xy+1)$\\

punto critico (0,0)\\$\frac{\delta^2 f}{\delta x^2}=2y+2x+2y$ evaluada en (0,0)=0 \\$\frac{\delta^2 f}{\delta y^2}= 2x+2y+2x$ evaluada en (0,0)=0\\$\frac{\delta^2 f}{\delta x \delta y}=2+2=0$\\

$D=(0)(0)=0

8/15/2019 comandos tarea 2

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$(-1 ,\frac{\pi}{4})$\\f=$\frac{\pi}{4}-1sen(\frac{\pi}{4})$\\f=$\frac{\pi}{4}-\frac{1}{\surd 2}$\\(-1,0) f=0+-1sen0 f=0\\Punto de inflexion\\\\\\

5)g=x+y+z=120 f=xy+xz+yz\\

$\bigtriangledown f \lambda \bigtriangledown g$\\

$\frac{\delta f}{\delta x}=y+z$\\

$\frac{\delta f}{\delta y}=x+z$\\$\frac{\delta f}{\delta z}=x+y$\\

$\frac{\delta g}{\delta z}=1$\\$\frac{\delta g}{\delta y}=1$\\$\frac{\delta g}{\delta z}=1$\\

y+z=\lambda\\x+z=\lambda\\x+y=\lambda\\

y+z=x+y= z=x\\y+z=x+z y=z\\

x=y=z\\

sustituyendo en g\\x+y+z=120 3x=120\\x+x+x=120 x=40\\

\end{document}