(1 point) Suppose f ( x , y )= x y (110 x 2 y ) f(x,y)=xy(110x2y) . f ( x , y )
ID: 2856456 • Letter: #
Question
(1 point) Suppose f(x,y)=xy(110x2y) f(x,y)=xy(110x2y) .
f(x,y) f(x,y) has 4 critical points. List them in increasing lexographic order. By that we mean that (x, y) comes before (z, w) if x<z x<z or if x=z x=z and y<w y<w . Also, describe the type of critical point by typing MA if it is a local maximum, MI if it is a local minimim, and S if it is a saddle point.
First point (
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Explanation / Answer
f(x,y)=xy(110x2y)
f(x,y)=xy10x2y2xy2
for critical points
fx=0 ,fy=0
fx=y20xy2y2=0
y(1-20x-2y)=0
y=0 ,1-20x-2y=0
fy=x10x24xy=0
x(1-10x-4y)=0
x=0, 1-10x-4y=0
(x,y)=(0,0) is one critical point
1-20x-2y=0 ,x=0
=>1-0-2y=0
=>y=1/2
(x,y)=(0,1/2) is critical point
y=0,1-10x-4y=0
1-10x-0=0
x=1/10
(x,y)=(1/10 ,0) is critical point
1-20x-2y=0 ,1-10x-4y=0
2(1-20x-2y)-(1-10x-4y)=0
2-40x-4y-1+10x+4y=0
1-30x=0
x=1/30
1-10x-4y=0
1-10(1/30)-4y=0
4y=2/3
y=1/6
(x,y)=(1/30,1/6) is critical point
all critical points are (x,y)=(0,0),(0,1/2),(1/30,1/6),(1/10 ,0)
fxx=-20y,fyy=-4x ,fxy =120x4y
D=fxxfyy-fxy2
D=(-20y)(-4x)-(120x4y)2
D=(80xy)-(120x4y)2
at (x,y)=(0,0),D=(0)-(1)2=-1<0 =>saddle point
at (x,y)=(0,1/2),D=(0)-(1-2)2=-1<0 =>saddle point
at (x,y)=(1/30,1/6),D=(80*(1/30)(1/6))-(1-(20/30-(4/6)))2=(4/9)-(1/9)=1/3>0 ,fxx=-20(1/6)<0 =>local maximum
at (x,y)=(1/10,0),D=(0)--(120(1/10)0)2=-1<0 =>saddle point
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