Critical points and extremum of $f(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2$
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Let $f:BbbR^2to BbbR$ be a function defined by
beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?
My work:
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have
beginalignx=frac12(3x^2+2xy)endalign
Substituting into $2$, we get
beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign
I don't know where to go from here. Can someone please, help?
calculus multivariable-calculus optimization linear-programming nonlinear-optimization
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up vote
2
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Let $f:BbbR^2to BbbR$ be a function defined by
beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?
My work:
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have
beginalignx=frac12(3x^2+2xy)endalign
Substituting into $2$, we get
beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign
I don't know where to go from here. Can someone please, help?
calculus multivariable-calculus optimization linear-programming nonlinear-optimization
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $f:BbbR^2to BbbR$ be a function defined by
beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?
My work:
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have
beginalignx=frac12(3x^2+2xy)endalign
Substituting into $2$, we get
beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign
I don't know where to go from here. Can someone please, help?
calculus multivariable-calculus optimization linear-programming nonlinear-optimization
Let $f:BbbR^2to BbbR$ be a function defined by
beginalignf(x,y)=y^2-x^2+x^3+x^2y+fracy^33;;forall;(x,y)inBbbR^2endalign
$i.$ Compute the critical points of $f$
$ii.$ Does $f$ have an extremum?
My work:
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have
beginalignx=frac12(3x^2+2xy)endalign
Substituting into $2$, we get
beginalign0=y(8+12x^3)+y^2(4x^2+4)+9x^4endalign
beginaligny=frac-2-3 x^3pmsqrt4+12 x^3-9 x^42 left(1+x^2right)endalign
I don't know where to go from here. Can someone please, help?
calculus multivariable-calculus optimization linear-programming nonlinear-optimization
edited Aug 23 at 4:36
asked Aug 23 at 4:29
Mike
75615
75615
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1 Answer
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beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have (1) implies
beginalignx(3x+2y-2)=0endalign
So there are two cases $x=0$ or ... continue from there.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have (1) implies
beginalignx(3x+2y-2)=0endalign
So there are two cases $x=0$ or ... continue from there.
add a comment |Â
up vote
3
down vote
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have (1) implies
beginalignx(3x+2y-2)=0endalign
So there are two cases $x=0$ or ... continue from there.
add a comment |Â
up vote
3
down vote
up vote
3
down vote
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have (1) implies
beginalignx(3x+2y-2)=0endalign
So there are two cases $x=0$ or ... continue from there.
beginalignfracpartial fpartial x=-2x+3x^2+2xyqquad (1)endalign
beginalignfracpartial fpartial y=2y+x^2+y^2qquad (2)endalign
At beginalignfracpartial fpartial x=fracpartial fpartial y=0endalign
we have (1) implies
beginalignx(3x+2y-2)=0endalign
So there are two cases $x=0$ or ... continue from there.
answered Aug 23 at 4:39
Andrew Allen
1197
1197
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