Odrediti lokalne ekstreme funkcije [inlmath]F(x,y)=\frac{1}{x}+\frac{1}{y}[/inlmath] pri uslovu [inlmath]\frac{1}{x^2}+\frac{1}{y^2}=1[/inlmath]
Evo sta sam ja odradila:
[dispmath]F(x,y,\lambda)=\frac{1}{x}+\frac{1}{y}+\lambda\left(\frac{1}{x^2}+\frac{1}{y^2}-1\right)\\
F'x=-\frac{1}{x^2}+\lambda\frac{-2}{x^3}\\
F'y=-\frac{1}{y^2}+\lambda\frac{-2}{y^3}\\
F'\lambda=\frac{1}{x^2}+\frac{1}{y^2}-1=0\\
-\frac{1}{x^2}+\lambda\frac{-2}{x^3}=0\\
-\frac{x^3}{x^2}-2\lambda=0,\;x=-2\lambda\\
y=-2\lambda\\
\frac{1}{(-2\lambda)^2}+\frac{1}{(-2\lambda)^2}=1,\;\lambda=\pm\frac{1}{\sqrt2}\\
F''x=\frac{2}{x^3}+\lambda\frac{6}{x^4}\\
F''y=\frac{2}{y^3}+\lambda\frac{6}{y^4}\\
F''xy=0\\
\lambda=-\frac{1}{\sqrt2},\;M_1\left(\frac{2}{\sqrt2},\frac{2}{\sqrt2}\right)\\
\mathrm d^2F=\left(\frac{2}{x^3}+\lambda\frac{6}{x^4}\right)\mathrm d^2x+0+\left(\frac{2}{y^3}+\lambda\frac{6}{y^4}\right)\mathrm d^2y[/dispmath]
Dovde sam stala, ne znam sta sa ovim da radim da bih odredila da li je vece ili manje od [inlmath]0[/inlmath]...