help find critical points of multivariable function?

Question

Finf all critical points and classify as local max/min/saddle:

f(x,y) =x^3 + xy - y^3.

I got:

f(0,0) = 0 ; saddle
f(1,-3) = 25 ; local min
f(3,1) = 29 ; saddle

am I missing any?

Answer 1

∂f(x,y)/∂x = 3x² + y = 0
∂f(x,y)/∂y = x-3y² = 0

So we add the equations equal and get

3x² - 3y² + x + y = 0

3(x - y)(x + y) + (x + y) = 0

(x + y) [3(x - y) + 1] = 0

So you have y = -x or y = 1/3 + x

Go back to ∂f(x,y)/∂x and substitute y=-x

3x² + y = 0
3x² - x = 0
x(3x - 1) = 0 so x=0 or x=1/3

So we have (0,0) and (1/3,-1/3) for critical points.

Now substitute y=1/3 + x

3x² + y = 0
3x² + 1/3 + x = 0
9x² + 3x + 1 = 0

This has no real solutions.

So the only critical points are (0,0) and (1/3,-1/3)

Your points f(1,-3) and f(3,1) don't qualify as critical points because only one of the partial derivatives are satisfied.

To find out what kind of points these are we need the second partial derivatives

∂²f(x,y)/∂x² = 6x
∂²f(x,y)/∂y² = -6y

(0,0) - both partials are 0 so this is a saddle point
(1/3,-1/3) - both partials are positive so it is a relative minimum.

These are the only critical points I can find.

Answer 2

Multivariable Functions

Answer 3

fx=3x^2+y =0
fy= x-3y^2 =0
x=3y^2
so 27y^4+y=0 so y =0 and 27 y^3=-1 so y =-1/3
so the points are
(0,0) and (1/3,-1/3)
fxx= 6x
fyy=-6y
fxy= 1
so H= -1 at (0,0) saddle
at(1/3,-1/3)) H = 36/9-1=3>0 fx>0 local minimum

Answer 4

You're missing a "hole", or a minimum, at x = 1/3, y = -1/3. I don't know what you're talking about with f(1,-3), f(3,1), they don't appear to be either saddles or max/min. f(0,0) is a bona fide saddle point.

Answer 5

diff(4*x^2*y-y^4-8*ln(x), x) = 8*x*y - 8/x diff(8*x*y-8/x, x) = 8*y + 8/x^2 solve(8*x*y-8/x, x) x1 = 1/sqrt(y), x2 = - 1/sqrt(y) solve(8*y+8/x^2, x) x2 = - 1/sqrt(-y), x3 = 1/sqrt(-y) diff(4*x^2*y-y^4-8*ln(x), y) = 4*x^2 - 4*y^3 diff(4*x^2-4*y^3, y) = - 12*y^2 solve(4*x^2-4*y^3, y) y1 = x^(2/3), y2 = - (1/2)*x^(2/3) + (1/2*i)*sqrt(3)*x^(2/3), y3 = - (1/2)*x^(2/3) - (1/2*i)*sqrt(3)*x^(2/3) solve( - 12*y^2) y4 = 0, y5 = 0 solve(4.*x^2*y-y^4-8.*ln(x), x) x = exp( - 0.5*LambertW( - y*exp( - 0.25*y^4)) - 0.125*y^4) solve(4.*x^2*y-y^4-8.*ln(x), y) y = RootOf(- 4*x^2*_Z + _Z^4 + 8*ln(x))