Question

2. Să se determine extremele funcției
$z = {x}^{3} + {y}^{2} - 3x + 2y$

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Answer 1

Răspuns:

$\frac{\partial z}{\partial x}=3x^2-3=0, \; \frac{\partial z}{\partial y}=2y+2=0$. Rezulta $x^2=1$ deci $x_1=1,x_2=-1$. Din 2y+2=0 rezulta y=-1.

Avem doua puncte critice: (1,-1), (-1,-1).

$\frac{\partial^2 z}{\partial x^2}=6x, \frac{\partial^2 z}{\partial xy}=0, \frac{\partial^2 z}{\partial y^2}=2$

Matricea Hessiana este:

$H_f(x,y)=\begin{pmatrix} \frac{\partial^2 z}{\partial x^2} & \frac{\partial^2 z}{\partial xy} \\ \frac{\partial^2 z}{\partial xy} &\frac{\partial^2 z}{\partial y^2}\end{pmatrix}=\begin{pmatrix} 6x & 0 \\ 0 & 2 \end{pmatrix}$.

$H_f(1,-1)=\begin{pmatrix} 6 & 0 \\ 0 & 2 \end{pmatrix}$.

$\Delta_1=6>0,\;\Delta_2=12>0$ deci (1,-1) este punct de minim local.

$H_f(-1,-1)=\begin{pmatrix} -6 & 0 \\ 0 & 2\end{pmatrix}$.

$\Delta_1=-6<0,\;\Delta_2=-12<0$ deci (-1,-1) nu este punct de extrem.