Răspuns:
f(x,y)=2x³+2y³-54x-150y+3
a)
Prima derivata df(x,y)/dx=6x²-54 se considera variabila x si y constant
df(x,y)/dy=6y²-150
a 2-a derivata
d²f/dx²=12x
d²f/dy=12y
d²f/dx·dy=0
d²f/dydx=0
b)Punctul critic M(xo,yo)
df/dx=0
6x²-54=0
x²=9
x=±3
df/dy=0
6y²-150=0
y²=25
y=±5
M1( -3,-5)
M2=(-3,+5)
M3=(3,-5)
M4(3,5)
Revin imediat
Calculam A,B,C
A=d²f/dx²(M1)=12x=12·*-3)= -36
B=d2f(M1)/dxdy=0
C=d2f(M1)/d²y=12yo=12*(-5)= -60
B²-AC=0-(-36)*(-60)= -216<0
A= -36 => M1 este punct de maxim
M2(-3,5)
A=d²f(M2)/d²x=12x=12*(-3)=-36
B=d²f(M2)/dxdy=0
C=d²f(M2)/dy²=12y=12*5=60
B²-AC=0-4*(-36)*60=144*60>0 Punctul M2 nu este nici punct de minim nici punct de Maxim
M3(3,-5)
A=d²f(M3)/d²x=12x=12*3=36
B=0
C=d²f(M3)/d²y=12y=12*(-5)=-60
B²-AC=0-36*(-60)=2160>0 M3 nu e punct de extrem
M4(3,5)
A=12x*3=36
B=0
C=12*5-60
B²-4AC=0-36*60= -2160<0
A=36>0+>
M4 punct de minim
Explicație pas cu pas:
