== Exercise Sheet 7 - Extrene Values & Optimization

=== Exercise 1 @Exercise[7, 1]
#block(
  fill: luma(230),
  inset: 8pt,
  radius: 4pt,
[
	Investigate the local extrema of the functions
	#set enum(numbering: "(a)")
	+ $ f(x,y) = 7-x^2-4x-y^2+2y $
	+ $ f(x,y) = x^4 + 3y^6 $
	+ $ f(x,y) = e^(-x^2-y^2) $
])
	#set enum(numbering: "(a)")
+ $
	nabla f(x,y) &= (-2x-4 , -2y+2) \
	nabla f(x,y) &=^! (0,0) "Notwendiges Kriterium" \
	upright("I"): -2x-4 &= 0 \
	"II"		: -2y+2 &= 0 \
	=> x = -2 space&space y = 1 \
	H(-2,1) &= mat(
		-2 , 0 ;
		0 , -2 ;
	) \
	det(H) "negativ" &=> "Lokales Maximum" \
	=> "Minimum in" (-2, 1): f(-2,1) &= 7-4+8-1+2 = 12 checkmark \
"Lösung mal wieder falsch... Kek"
$
+ $
	nabla f(x,y) &= ( 4x^3 , 18y^5 ) \
	nabla f(x,y) &=^! (0,0) "Notwendiges Kriterium" \
	upright("I")	: 4x^3 &= 0 \
	upright("II")	: 18y^5 &= 0 \
	=> x &= y = 0 \
	H(x,y) &= mat(
		4x^3 , 0 ;
		0 , 90y^4 ;
	) \
	=> H(0,0) &= mat(0,0;0,0) "indeterminiert" \
	f &"ist immer positiv, daher muss das extrema ein Minimum sein." \
	=> P(0,0) &"ist ein lokales Minimum mit" f(0,0) = 0
$
+ $
	nabla f(x,y) &= (e^(-x^2-y^2) dot (-2 x) , e^(-x^2-y^2) dot (-2 y) ) \
	=> x &= y = 0 "(Notwendige Bedingung)" \
	=> H(x,y) &= mat(
		2x^2-1 , 4x y ;
		4x y , 2y^2-1 ;
	) dot e^(-x^2-y^2) "Achtung, Produktregel"\
	H(0,0) &= mat(
		-1,0;
		0, -1
	) => "Maximum" \
	f(0,0) &= 1 checkmark

$
