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Loading Student:-MultivariateCalculus

$f\=x^3-y^3-3 x y\+4$$\stackrel{\text{assign}}{\to }$

Solution via task template

Tools≻Tasks≻Browse:

Calculus - Multivariate≻Optimization≻

Find critical points via first principles

$\frac{\partial }{\partial x} f\=0\,\frac{\partial }{\partial y} f\=0$

$3x^2-3y\=0\,-3y^2-3x\=0$

$\stackrel{\text{solve}}{\to }$

$\left\{x\=0\,y\=0\right\}\,\left\{x\=-1\,y\=1\right\}\,\left\{x\=\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\,y\=-\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\right\}\,\left\{x\=\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\,y\=-\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\right\}$

Alternate calculation of the critical points

$f$

$x^3-y^3-3xy\+4$

$\stackrel{\text{gradient}}{\to }$

$\stackrel{\text{to list}}{\to }$

$\left[3x^2-3y\,-3y^2-3x\right]$

$\stackrel{\text{equate to 0}}{\to }$

$\left[3x^2-3y\=0\,-3y^2-3x\=0\right]$

$\stackrel{\text{solve}}{\to }$

$\left\{x\=0\,y\=0\right\}\,\left\{x\=-1\,y\=1\right\}\,\left\{x\=\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\,y\=-\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\right\}\,\left\{x\=\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\,y\=-\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\right\}$

Second-Derivative test at $\left(-1\,1\right)$

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)\cdot \left(\frac{{\partial }^2}{{\partial }^{}{y}^2} f\right)-{\left(\frac{{\partial }^2}{\partial y\partial x} f\right)}^2}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=-1\,y\=1}$ = $27$$$

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=-1\,y\=1}$ = $-6$$$

Obtain $f\left(-1\,1\right)$ 

$\genfrac{}{}{0ex}{}{f}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=-1\,y\=1}$ = $5$

Second-Derivative test at $\left(0\,0\right)$ 

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)\cdot \left(\frac{{\partial }^2}{{\partial }^{}{y}^2} f\right)-{\left(\frac{{\partial }^2}{\partial y\partial x} f\right)}^2}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=0\,y\=0}$ = $-9$$$

Initialize

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{MultivariateCalculus}\right)\:$

$f≔\left(x\,y\right)\to x^3-y^3-3 x y\+4\:$

Obtain critical points

$\mathrm{solve}\left(\mathrm{Equate}\left(\mathrm{Gradient}\left(f\left(x\,y\right)\,\left[x\,y\right]\right)\,⟨0\,0⟩\right)\,\mathrm{explicit}\right)$

$\left\{x\=0\,y\=0\right\}\,\left\{x\=-1\,y\=1\right\}\,\left\{x\=\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\,y\=-\frac{1}{2}\+\frac{1}{2}\mathrm{I}\sqrt{3}\right\}\,\left\{x\=\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\,y\=-\frac{1}{2}-\frac{1}{2}\mathrm{I}\sqrt{3}\right\}$

Apply the SecondDerivativeTest command to the critical point

$\mathrm{SecondDerivativeTest}\left(f\left(x\,y\right)\,\left[x\,y\right]\=\left[\left[0\,0\right]\,\left[-1\,1\right]\right]\right)$

$\mathrm{LocalMin}\=\left[\right]\,\mathrm{LocalMax}\=\left[\left[-1\,1\right]\right]\,\mathrm{Saddle}\=\left[\left[0\,0\right]\right]$

Apply the second-derivative test from first principles

The differential operator D, applied to a function, returns a function. Hence, $T\=f_{\mathrm{xx}}{f}_{\mathrm{yy}}-f_{\mathrm{xy}}^2$ is a function evaluated at the one critical point.

$$\begin{gathered} T≔\left({\mathrm{D}}_{1\,1}\left(f\right)\cdot {\mathrm{D}}_{2\,2}\left(f\right)-{\mathrm{D}}_{1\,2}{\left(f\right)}^2\right)\left(-1\,1\right)\; \\ {\mathrm{D}}_{1\,1}\left(f\right)\left(-1\,1\right) \end{gathered}$$

$27$

$T≔\left({\mathrm{D}}_{1\,1}\left(f\right)\cdot {\mathrm{D}}_{2\,2}\left(f\right)-{\mathrm{D}}_{1\,2}{\left(f\right)}^2\right)\left(0\,0\right)$

$-9$

Apply Sylvester's criterion at $\left(-1\,1\right)$

$H≔\mathrm{SecondDerivativeTest}\left(f\left(x\,y\right)\,\left[x\,y\right]\=\left[-1\,1\right]\,\mathrm{output}\=\mathrm{hessian}\right)$ = $$

Generate the sequence $1\,Q_1\,\dots \,Q_n$, where the $Q_k$ are the principal minors.

$1\, \mathrm{seq}\left(\mathrm{LinearAlgebra}:-\mathrm{Determinant}\left(H\left(1..k\,1..k\right)\right)\,k\=1..2\right)$

$1\,-6\,27$

$\mathrm{Matrix}\left(2\,2\,\left(i\,j\right)\to {\mathrm{D}}_{i\,j}\left(f\right)\left(-1\,1\right)\right)$ = $$

Apply Sylvester's criterion at $\left(0\,0\right)$

$H≔\mathrm{SecondDerivativeTest}\left(f\left(x\,y\right)\,\left[x\,y\right]\=\left[0\,0\right]\,\mathrm{output}\=\mathrm{hessian}\right)$ = $$

Generate the sequence $1\,Q_1\,\dots \,Q_n$, where the $Q_k$ are the principal minors.

$1\, \mathrm{seq}\left(\mathrm{LinearAlgebra}:-\mathrm{Determinant}\left(H\left(1..k\,1..k\right)\right)\,k\=1..2\right)$

$1\,0\,-9$

$\mathrm{Matrix}\left(2\,2\,\left(i\,j\right)\to {\mathrm{D}}_{i\,j}\left(f\right)\left(0\,0\right)\right)$ = $$