Odpowiedź:
Maksimum lokalne:
[tex]$f\Big(-\frac{1}{2} ,\frac{1}{4} ,-\frac{1}{4} \Big)=\frac{1}{2}[/tex]
Minimum lokalne:
[tex]$f\Big(\frac{1}{2} ,\frac{1}{4} ,\frac{1}{4} \Big)=-\frac{1}{2}[/tex]
Szczegółowe wyjaśnienie:
Funkcja:
[tex]$f(x,y,z)=\frac{x^{3}}{y} -3x+z+\frac{y^{2}}{z}[/tex]
Dziedzina:
[tex]D=\Big\{(x,y,z) \in \mathbb{R}^{3} : y \neq 0 \wedge z \neq 0 \Big\}[/tex]
Pochodne cząstkowe rzędu pierwszego:
[tex]$\frac{\partial f}{\partial x} =\frac{3x^{2}}{y} -3[/tex]
[tex]$\frac{\partial f}{\partial y} =-\frac{x^{3}}{y^{2}} +\frac{2y}{z}[/tex]
[tex]$\frac{\partial f}{\partial z}=1-\frac{y^{2}}{z^{2}}[/tex]
Układ równań:
[tex]$\left\{\begin{array}{ccc}\frac{3x^{2}}{y}-3=0 \\-\frac{x^{3}}{y^{2}}+\frac{2y}{z} =0 \\1-\frac{y^{2}}{z^{2}}=0 \end{array}\right[/tex]
Z trzeciego równania mamy:
[tex]$z^{2}-y^{2}=0 \iff (z-y)(z+y)=0 \iff y = z \vee y = -z[/tex]
Dla [tex]y=-z[/tex] mamy:
[tex]$\left \{ {{-\frac{3x^{2}}{z}-3=0 } \atop {-\frac{x^{3}}{z^{2}}-2=0 }} \right.[/tex]
[tex]$\left \{ {{x^{2}+z=0} \atop {x^{3}+2z^{2}=0}} \right.[/tex]
Z pierwszego [tex]z=-x^{2}[/tex] :
[tex]$x^{3}+2x^{4}=0\\[/tex]
[tex]$x^{3}(2x+1)=0 \iff x=0 \vee x=-\frac{1}{2}[/tex]
Zauważmy, że gdy [tex]x=0[/tex], to [tex]z=0[/tex], a zatem jest to sprzeczne z dziedziną.
Mamy następujące rozwiązanie:
[tex]$P_{1}=\Big(-\frac{1}{2} ,\frac{1}{4} ,-\frac{1}{4} \Big)[/tex]
Dla [tex]y=z[/tex] mamy:
[tex]$\left \{ {{\frac{3x^{2}}{z}-3=0 } \atop {-\frac{x^{3}}{z^{2}}+2=0 }} \right.[/tex]
[tex]$\left \{ {{x^{2}-z=0} \atop {-x^{3}+2z^{2}=0}} \right.[/tex]
Z pierwszego [tex]z=x^{2}[/tex] :
[tex]$-x^{3}+2x^{4}=0[/tex]
[tex]$x^{3}(2x-1)=0 \iff x =0 \vee x=\frac{1}{2}[/tex]
Stąd mamy drugie rozwiązanie:
[tex]$P_{2}=\Big(\frac{1}{2} ,\frac{1}{4} ,\frac{1}{4} \Big)[/tex]
Pochodne cząstkowe rzędu drugiego:
[tex]$\frac{\partial ^{2}f}{\partial x^{2}} =\frac{6x}{y}[/tex]
[tex]$\frac{\partial ^{2}f}{\partialy \partial y \partial x} =-\frac{3x^{2}}{y^{2}}[/tex]
[tex]$\frac{\partial ^{2}f}{\partial z \partial x}=0[/tex]
[tex]$\frac{\partial ^{2}f}{\partial x \partial y}=-\frac{3x^{2}}{y^{2}}[/tex]
[tex]$\frac{\partial ^{2}f}{\partial y^{2}}=\frac{2x^{3}}{y^{3}} +\frac{2}{z}[/tex]
[tex]$\frac{\partial ^{2}f}{\partial z \partial y}=-\frac{2y}{z^{2}}[/tex]
[tex]$\frac{\partial ^{2}f}{\partial x \partial z}=0[/tex]
[tex]$\frac{\partial ^{2}f}{\partial y \partial z}=-\frac{2y}{z^{2}}[/tex]
[tex]$\frac{\partial ^{2}f}{\partial z^{2}}=\frac{2y^{2}}{z^{3}}[/tex]
Wyznaczniki będą następujące:
[tex]$W_{3} =\left|\begin{array}{ccc}\frac{\partial ^{2}f}{\partial x^{2}}&\frac{\partial ^{2}f}{\partialy \partial y \partial x}&\frac{\partial ^{2}f}{\partial z \partial x}\\\frac{\partial ^{2}f}{\partial x \partial y}&\frac{\partial ^{2}f}{\partial y^{2}}&\frac{\partial ^{2}f}{\partial z \partial y}\\\frac{\partial ^{2}f}{\partial x \partial z}&\frac{\partial ^{2}f}{\partial y \partial z}&\frac{\partial ^{2}f}{\partial z^{2}}\end{array}\right|[/tex]
[tex]$W_{2}=\left|\begin{array}{ccc}\frac{\partial ^{2}f}{\partial x^{2}} &\frac{\partial ^{2}f}{\partialy \partial y \partial x}\\\frac{\partial ^{2}f}{\partialy \partial x \partial y}&\frac{\partial ^{2}f}{\partialy \partial y^{2}}\end{array}\right|[/tex]
[tex]$W_{1}=\Big| \frac{\partial ^{2}f}{\partial x^{2}}\Big|[/tex]
Dla punktu [tex]P_{1}[/tex] są one nastepujące:
[tex]W_{3}(P_{1})=\left|\begin{array}{ccc}-12&-12&0\\-12&-24&-8\\0&-8&-8\end{array}\right|=-384 < 0[/tex]
[tex]W_{2}(P_{1})=\left|\begin{array}{ccc}-12&-12\\-12&-24\end{array}\right|=144 > 0[/tex]
[tex]W_{1}(P_{1})=-12 < 0[/tex]
Mamy więc:
[tex]W_{1}(P_{1}) < 0 \ , \ W_{2}(P_{1}) > 0 \ \text{i} \ W_{3}(P_{1}) < 0[/tex]
Zatem w punkcie [tex]P_{1}[/tex] mamy maksimum lokalne równe:
[tex]$f\Big(-\frac{1}{2} ,\frac{1}{4} ,-\frac{1}{4} \Big)=\frac{1}{2}[/tex]
Dla punktu [tex]P_{2}[/tex] mamy:
[tex]W_{3}(P_{2})=\left|\begin{array}{ccc}12&-12&0\\-12&24&-8\\0&-8&8\end{array}\right|=384 > 0[/tex]
[tex]W_{2}(P_{2})=\left|\begin{array}{ccc}12&-12\\-12&24\end{array}\right|=144 > 0[/tex]
[tex]W_{3}(P_{2})=12 > 0[/tex]
Zatem mamy:
[tex]W_{1}(P_{1}) > 0 \ , \ W_{2}(P_{1}) > 0 \ \text{i} \ W_{3}(P_{1}) > 0[/tex]
Zatem w punkcie [tex]P_{2}[/tex] mamy minimum lokalne równe:
[tex]$f\Big(\frac{1}{2} ,\frac{1}{4} ,\frac{1}{4} \Big)=-\frac{1}{2}[/tex]
