Figure 4.8.1(a)   The surface defined by $f$ 

Initialize

Loading Student:-MultivariateCalculus

$f\=7-3 x^2-5 y^2\+6 x-9 y\+4$$\stackrel{\text{assign}}{\to }$

Solution via task template

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Calculus - Multivariate≻Optimization≻

Find critical points via first principles

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

$-6x\+6\=0\,-10y-9\=0$

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

$\left\{x\=1\,y\=-\frac{9}{10}\right\}$

Alternate calculation of the critical points

$f$

$-3x^2-5y^2\+6x-9y\+11$

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

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

$\left[-6x\+6\,-10y-9\right]$

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

$\left[-6x\+6\=0\,-10y-9\=0\right]$

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

$\left\{x\=1\,y\=-\frac{9}{10}\right\}$

Second-Derivative test

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)\cdot \left(\frac{{\partial }^2}{{\partial }^{}{y}^2} f\right)-{\left(\frac{{\partial }^2}{\partial y\partial x} f\right)}^2}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=1\,y\=-9\/10}$ = $60$$$

$\genfrac{}{}{0ex}{}{\left(\frac{{\partial }^2}{{\partial }^{}{x}^2} f\right)}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=1\,y\=-9\/10}$ = $-6$$$

Obtain $f\left(1\,-9\/10\right)$ 

$\genfrac{}{}{0ex}{}{f}{\phantom{x\=a}}|\genfrac{}{}{0ex}{}{\phantom{\mathrm{f(x)}}}{x\=1\,y\=-9\/10}$ = $\frac{361}{20}$$\stackrel{\text{at 5 digits}}{\to }$$18.050$$$

Initialize

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

$f≔\left(x\,y\right)\to 7-3 x^2-5 y^2\+6 x-9 y\+4\:$

Obtain critical points

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

$\left\{x\=1\,y\=-\frac{9}{10}\right\}$

Apply the SecondDerivativeTest command to the critical point

$\mathrm{SecondDerivativeTest}\left(f\left(x\,y\right)\,\left[x\,y\right]\=\left[1\,-9\/10\right]\right)$

$\mathrm{LocalMin}\=\left[\right]\,\mathrm{LocalMax}\=\left[\left[1\,-\frac{9}{10}\right]\right]\,\mathrm{Saddle}\=\left[\right]$

Apply the second-derivative test from first principles

$$\begin{gathered} T≔\left({\mathrm{D}}_{1\,1}\left(f\right)\cdot {\mathrm{D}}_{2\,2}\left(f\right)-{\mathrm{D}}_{1\,2}{\left(f\right)}^2\right)\left(1\,-9\/10\right)\; \\ {\mathrm{D}}_{1\,1}\left(f\right)\left(1\,-9\/10\right) \end{gathered}$$

$60$

Apply Sylvester's criterion

$H≔\mathrm{SecondDerivativeTest}\left(f\left(x\,y\right)\,\left[x\,y\right]\=\left[1\,-9\/10\right]\,\mathrm{output}\=\mathrm{hessian}\right)$ = $$

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

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

$1\,-6\,60$

$\mathrm{Matrix}\left(2\,2\,\left(i\,j\right)\to {\mathrm{D}}_{i\,j}\left(f\right)\left(1\,-9\/10\right)\right)$ = $$