Lokalni ekstremi realne funkcije dvije varijable

verzija: SageMath 9.4

In [1]:
from sage.plot.plot3d.shapes import *
In [2]:
%display latex
In [3]:
var('y z k')
Out[3]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(y, z, k\right)\]

1. zadatak

Odredite lokalne ekstreme i sedlaste točke funkcije $f(x,y)=e^{x-y}\big(x^2-2y^2\big).$

Rješenje

In [4]:
f(x,y)=e^(x-y)*(x^2-2*y^2)

Gradijent funkcije $f$

In [5]:
grad_f=f.diff(); grad_f
Out[5]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left( x, y \right) \ {\mapsto} \ \left({\left(x^{2} - 2 \, y^{2}\right)} e^{\left(x - y\right)} + 2 \, x e^{\left(x - y\right)},\,-{\left(x^{2} - 2 \, y^{2}\right)} e^{\left(x - y\right)} - 4 \, y e^{\left(x - y\right)}\right)\]
In [6]:
grad_f.apply_map(lambda t: t.factor())
Out[6]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left({\left(x^{2} - 2 \, y^{2} + 2 \, x\right)} e^{\left(x - y\right)},\,-{\left(x^{2} - 2 \, y^{2} + 4 \, y\right)} e^{\left(x - y\right)}\right)\]

Stacionarne točke: $(0,0),\ (-4,-2)$

In [7]:
solve([grad_f[0]==0,grad_f[1]==0],[x,y])
Out[7]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left[\left[x = \left(-4\right), y = \left(-2\right)\right], \left[x = 0, y = 0\right]\right]\]

Hesseova matrica

In [8]:
hess_f=f.diff(2); hess_f
Out[8]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} \left( x, y \right) \ {\mapsto} \ {\left(x^{2} - 2 \, y^{2}\right)} e^{\left(x - y\right)} + 4 \, x e^{\left(x - y\right)} + 2 \, e^{\left(x - y\right)} & \left( x, y \right) \ {\mapsto} \ -{\left(x^{2} - 2 \, y^{2}\right)} e^{\left(x - y\right)} - 2 \, x e^{\left(x - y\right)} - 4 \, y e^{\left(x - y\right)} \\ \left( x, y \right) \ {\mapsto} \ -{\left(x^{2} - 2 \, y^{2}\right)} e^{\left(x - y\right)} - 2 \, x e^{\left(x - y\right)} - 4 \, y e^{\left(x - y\right)} & \left( x, y \right) \ {\mapsto} \ {\left(x^{2} - 2 \, y^{2}\right)} e^{\left(x - y\right)} + 8 \, y e^{\left(x - y\right)} - 4 \, e^{\left(x - y\right)} \end{array}\right)\]
In [9]:
hess_f = hess_f.apply_map(lambda t: t.factor()); hess_f
Out[9]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} {\left(x^{2} - 2 \, y^{2} + 4 \, x + 2\right)} e^{\left(x - y\right)} & -{\left(x^{2} - 2 \, y^{2} + 2 \, x + 4 \, y\right)} e^{\left(x - y\right)} \\ -{\left(x^{2} - 2 \, y^{2} + 2 \, x + 4 \, y\right)} e^{\left(x - y\right)} & {\left(x^{2} - 2 \, y^{2} + 8 \, y - 4\right)} e^{\left(x - y\right)} \end{array}\right)\]

Točka $(0,0)$ je sedlasta točka funkcije $f$.

In [10]:
hess_f(x=0,y=0)
Out[10]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} 2 & 0 \\ 0 & -4 \end{array}\right)\]
In [11]:
hess_f(x=0,y=0).det()
Out[11]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}-8\]

U točki $(-4,-2)$ funkcija $f$ ima lokalni maksimum koji iznosi $8e^{-2}$.

In [12]:
hess_f(x=-4,y=-2)
Out[12]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rr} -6 \, e^{\left(-2\right)} & 8 \, e^{\left(-2\right)} \\ 8 \, e^{\left(-2\right)} & -12 \, e^{\left(-2\right)} \end{array}\right)\]
In [13]:
hess_f(x=-4,y=-2).det()
Out[13]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}8 \, e^{\left(-4\right)}\]
In [14]:
f(-4,-2)
Out[14]:
\[\newcommand{\Bold}[1]{\mathbf{#1}}8 \, e^{\left(-2\right)}\]

Na slici je prikazan graf funkcije $f$ i istaknuta je svjetloplava sedlasta točka i žuta točka lokalnog maksimuma.

In [15]:
ploha1=implicit_plot3d(f(x,y)-z,(x,-5,5),(y,-5,5),(z,-7,7),color='pink',plot_points=60)
tf1=Sphere(0.1,color='yellow').scale(1,1,1.1).translate(-4,-2,f(-4,-2))
tf2=Sphere(0.1,color='cyan').scale(1,1,1.1).translate(0,0,f(0,0))
(ploha1+tf1+tf2).show(viewer='threejs',online=True,aspect_ratio=[1,1,0.8])