Initialize

Loading Student:-MultivariateCalculus

$f\=2 x y-3 x^4-5 y^4\+2-y\+7$$\stackrel{\text{assign}}{\to }$

Find critical point via first principles

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

$-12x^3+2y=0,-20y^3+2x-1=0$

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

$\left\{x=\mathrm{-0.4222805036}\,y=\mathrm{-0.4518084433}\right\}$

$\stackrel{\text{assign to a name}}{\to }$

$C$

Assign $X$ and $Y$ the values of $x$ and $y$ at the critical point

$\genfrac{}{}{0ex}{}{x}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{C}$ = $\mathrm{-0.4222805036}$$\stackrel{\text{assign to a name}}{\to }$$X$

$\genfrac{}{}{0ex}{}{y}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{C}$ = $\mathrm{-0.4518084433}$$\stackrel{\text{assign to a name}}{\to }$$Y$$$

Alternate calculation of the critical point

$f$

$-3x^4-5y^4+2xy-y+9$

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

$\left[\begin{array}{c}-12x^3+2y\\ -20y^3+2x-1\end{array}\right]$

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

$\left[-12x^3+2y\,-20y^3+2x-1\right]$

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

$\left[-12x^3+2y=0\,-20y^3+2x-1=0\right]$

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

$\left\{x=\mathrm{-0.4222805036}\,y=\mathrm{-0.4518084433}\right\}$

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

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)\cdot \left(\frac{{\partial }^2}{{\partial }^{}{y}^2} f\right)-{\left(\frac{{\partial }^2}{\partial y\partial x} f\right)}^2}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{C}$ = $74.62569511$$$

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{C}$ = $\mathrm{-6.419549653}$$$

Obtain $f\left(X\,Y\right)$ 

$\genfrac{}{}{0ex}{}{f}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{C}$ = $9.529646229$

Initialize

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

$f≔\left(x\,y\right)\to 2 x y-3 x^4-5 y^4\+2-y\+7\:$

Obtain critical points

$$\begin{gathered} C≔\mathrm{fsolve}\left(\mathrm{Equate}\left(\mathrm{Gradient}\left(f\left(x\,y\right)\,\left[x\,y\right]\right)\,⟨0\,0⟩\right)\,\left\{x\,y\right\}\right)\; \\ X≔\mathrm{eval}\left(x\,C\right)\; \\ Y≔\mathrm{eval}\left(y\,C\right) \end{gathered}$$

$Y≔\mathrm{-0.4518084433}$

Apply the SecondDerivativeTest command to the critical point

$\mathrm{SecondDerivativeTest}\left(f\left(x\,y\right)\,\left[x\,y\right]\=\left[X\,Y\right]\right)$

$\mathrm{LocalMin}=\left[\right],\mathrm{LocalMax}=\left[\left[\mathrm{-0.4222805036}\,\mathrm{-0.4518084433}\right]\right],\mathrm{Saddle}=\left[\right]$

Apply the second-derivative test from first principles

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

$\mathrm{-6.419549653}$

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

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

$H≔\left[\begin{array}{cc}\mathrm{-6.419549653}& 2\\ 2& \mathrm{-12.24785216}\end{array}\right]$

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

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

$1,\mathrm{-6.419549653},74.62569508$

$\mathrm{Matrix}\left(2\,2\,\left(i\,j\right)\to {\mathrm{D}}_{i\,j}\left(f\right)\left(X\,Y\right)\right)$ = $\left[\begin{array}{cc}\mathrm{-6.419549653}& 2\\ 2& \mathrm{-12.24785216}\end{array}\right]$$$