Екстремум на фунцкия

[Гост]

Здравейте, имам да намеря локалния екстремум на тази фунцкия (10.156), като е даден и отговор.

[ammornil]

[tex]z=x^{2}+y^{2}+xy-3ax-3by \\ \begin{array}{lclcl} \quad \\ \frac{\normalsize{\partial{z}}}{\normalsize{\partial{x}}}=2x+y-3a & \quad & \frac{\normalsize{\partial{z}}}{\normalsize{\partial{y}}}=2y+x-3b \\ \quad \\ \frac{\normalsize{\partial^{2}{z}}}{\normalsize{\partial{x^{2}}}}=2 & \quad & \frac{\normalsize{\partial^{2}{z}}}{\normalsize{\partial{y^{2}}}}=2 & \quad & \frac{\normalsize{\partial^{2}{z}}}{\normalsize{\partial{x}\partial{y}}}=1 \\ \quad \end{array} \\ \frac{\partial^{2}{z}}{\partial{x^{2}}}\cdot{} \frac{\partial^{2}{z}}{\partial{y^{2}}}-\left(\frac{\partial^{2}{z}}{\partial{x}\partial{y}} \right)^{2}>0 \hspace{0.5em} \forall{(x;y)} \quad \Rightarrow \exists{z_{extr}} \\ \frac{\partial^{2}{z}}{\partial{x^{2}}}>0 \hspace{0.5em} \forall{(x;y)} \Rightarrow z_{extr}=z_{min} \rightarrow \begin{array}{|l} \frac{\partial{z}}{\partial{x}}=0 \\ \frac{\partial{z}}{\partial{y}}= 0 \end{array} \\ \begin{array}{|l} 2x+y-3a=0 \quad |\cdot{}(-2) \\ x+2y-3b=0 \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} y=3a-2x \\ -3x+6a-3b=0 \end{array} \quad \Leftrightarrow \\ \quad \Leftrightarrow \quad \begin{array}{|l} y=3a-2(2a-b) \\ x=2a-b \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} y=2b-a \\ x=2a-b \end{array} \\ z_{min}=z(2a-b;2b-a)=(2a-b)^{2}+(2b-a)^{2}+(2a-b)(2b-a)-3a(2a-b)-3b(2b-a) \\ z_{min}=\red{4a^{2}}\purple{-4ab}\blue{+b^{2}}\blue{+4b^{2}}\purple{-4ab}\red{+a^{2}}\purple{+4ab}\red{-2a^{2}}\blue{-2b^{2}}\purple{+ab}\red{-6a^{2}}\purple{+3ab}\blue{-6b^{2}}\purple{+3ab} \\ z_{min}=-3a^{2}-3b^{2}+3ab[/tex]

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[tex]a=1, b=1 \rightarrow Graph{(z)}\\[/tex]

Screenshot 2024-05-24 224122.png (95.44 KiB) Прегледано 355 пъти

[Гост]

[tex]z=x^{2}+y^{2}+xy-3ax-3by \\ \begin{array}{lclcl} \quad \\ \frac{\normalsize{\partial{z}}}{\normalsize{\partial{x}}}=2x+y-3a & \quad & \frac{\normalsize{\partial{z}}}{\normalsize{\partial{y}}}=2y+x-3b \\ \quad \\ \frac{\normalsize{\partial^{2}{z}}}{\normalsize{\partial{x^{2}}}}=2 & \quad & \frac{\normalsize{\partial^{2}{z}}}{\normalsize{\partial{y^{2}}}}=2 & \quad & \frac{\normalsize{\partial^{2}{z}}}{\normalsize{\partial{x}\partial{y}}}=1 \\ \quad \end{array} \\ \frac{\partial^{2}{z}}{\partial{x^{2}}}\cdot{} \frac{\partial^{2}{z}}{\partial{y^{2}}}-\left(\frac{\partial^{2}{z}}{\partial{x}\partial{y}} \right)^{2}>0 \hspace{0.5em} \forall{(x;y)} \quad \Rightarrow \exists{z_{extr}} \\ \frac{\partial^{2}{z}}{\partial{x^{2}}}>0 \hspace{0.5em} \forall{(x;y)} \Rightarrow z_{extr}=z_{min} \rightarrow \begin{array}{|l} \frac{\partial{z}}{\partial{x}}=0 \\ \frac{\partial{z}}{\partial{y}}= 0 \end{array} \\ \begin{array}{|l} 2x+y-3a=0 \quad |\cdot{}(-2) \\ x+2y-3b=0 \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} y=3a-2x \\ -3x+6a-3b=0 \end{array} \quad \Leftrightarrow \\ \quad \Leftrightarrow \quad \begin{array}{|l} y=3a-2(2a-b) \\ x=2a-b \end{array} \quad \Leftrightarrow \quad \begin{array}{|l} y=2b-a \\ x=2a-b \end{array} \\ z_{min}=z(2a-b;2b-a)=(2a-b)^{2}+(2b-a)^{2}+(2a-b)(2b-a)-3a(2a-b)-3b(2b-a) \\ z_{min}=\red{4a^{2}}\purple{-4ab}\blue{+b^{2}}\blue{+4b^{2}}\purple{-4ab}\red{+a^{2}}\purple{+4ab}\red{-2a^{2}}\blue{-2b^{2}}\purple{+ab}\red{-6a^{2}}\purple{+3ab}\blue{-6b^{2}}\purple{+3ab} \\ z_{min}=-3a^{2}-3b^{2}+3ab[/tex]

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[tex]a=1, b=1 \rightarrow Graph{(z)}\\[/tex]

Screenshot 2024-05-24 224122.png

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