Find the local maxima and minima for##f(x,y) = x^3-xy-x+xy^3-y^4##

In summary, to find the local maxima and minima of the function \( f(x,y) = x^3 - xy - x + xy^3 - y^4 \), we first compute the critical points by taking the partial derivatives \( f_x \) and \( f_y \), setting them equal to zero, and solving the resulting equations. This involves finding \( f_x = 3x^2 - y - 1 + y^3 = 0 \) and \( f_y = -x + 3xy^2 - 4y^3 = 0 \). After locating the critical points, we apply the second derivative test using the Hessian determinant to classify each critical point as a local maximum, local minimum,
  • #1
chwala
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Homework Statement
see attached.
Relevant Equations
##\nabla f = 0##
1701685037726.png


Ok i have,

##f_x= 3x^2-y-1+y^3##

##f_y = -x+3xy^2-4y^3##

##f_{xx} = 6x##

##f_{yy} = 6xy - 12y^2##

##f_{xy} = -1+3y^2##

looks like one needs software to solve this?

I can see the solutions from wolframalpha: local maxima to two decimal places as;

##(x,y) = (-0.67, 0.43)##

...but i am more interested in steps that lead to the given solution...
 
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  • #2
chwala said:
looks like one needs software to solve this?

Indeed. See
https://www.wolframalpha.com/input?i=3x^2+y^3-y=1+AND+4y^3=x(3y^2-1)

chwala said:
I can see the solutions from wolframalpha: local maxima to two decimal places as;

##(x,y) = (-0.67, 0.43)##

...but i am more interested in steps that lead to the given solution...

Look at the plot. This gives you an idea of how a numerical algorithm could work. Walk along the blue line until you cross the orange line and determine whether it is a local minimum, a local maximum, or an inflection point.
 
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  • #4
fresh_42 said:
Indeed. See
https://www.wolframalpha.com/input?i=3x^2+y^3-y=1+AND+4y^3=x(3y^2-1)
Look at the plot. This gives you an idea of how a numerical algorithm could work. Walk along the blue line until you cross the orange line and determine whether it is a local minimum, a local maximum, or an inflection point.
Thanks from the plot we have the point ##(-7.540, -5.595)## being an inflection point or can we say saddle point? then ##(0.471, -0.396)## being the local minimum... bringing me to the next question, do we have a global maximum and global minimum for this problem?
 
  • #5
chwala said:
Thanks from the plot we have the point ##(-7.540, -5.595)## being an inflection point or can we say saddle point?
Yes.
chwala said:
then ##(0.471, -0.396)## being the local minimum... bringing me to the next question, do we have a global maximum and global minimum for this problem?
Look at the links in @anuttarasammyak 's post #3.
 
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FAQ: Find the local maxima and minima for##f(x,y) = x^3-xy-x+xy^3-y^4##

What are the critical points of the function \( f(x,y) = x^3 - xy - x + xy^3 - y^4 \)?

To find the critical points, we need to set the partial derivatives of the function with respect to \( x \) and \( y \) to zero. The partial derivatives are:

\(\frac{\partial f}{\partial x} = 3x^2 - y - 1\)

\(\frac{\partial f}{\partial y} = -x + 3xy^2 - 4y^3\)

Setting these equations to zero, we solve for \( x \) and \( y \):\[ 3x^2 - y - 1 = 0 \]\[ -x + 3xy^2 - 4y^3 = 0 \]Solving these equations simultaneously will give the critical points.

How do you classify the critical points as local maxima, minima, or saddle points?

To classify the critical points, we use the second derivative test. We need to find the second partial derivatives:

\(\frac{\partial^2 f}{\partial x^2} = 6x\)

\(\frac{\partial^2 f}{\partial y^2} = 6xy - 12y^2\)

\(\frac{\partial^2 f}{\partial x \partial y} = -1 + 3y^2\)

We then evaluate the Hessian determinant \( D = f_{xx} f_{yy} - (f_{xy})^2 \) at each critical point. If \( D > 0 \) and \( f_{xx} > 0 \), we have a local minimum. If \( D > 0 \) and \( f_{xx} < 0 \), we have a local maximum. If \( D < 0 \), the point is a saddle point.

What are the explicit coordinates of the critical points for the function \( f(x,y) \)?

Solving the system of equations derived from the first derivatives:\[ 3x^2 - y - 1 = 0 \]\[ -x + 3xy^2 - 4y^3 = 0 \]we find the critical points. This typically involves solving a polynomial system, which may require numerical methods or further algebraic manipulation. The explicit coordinates of the critical points are the solutions to these equations.

Can the function \( f(x,y) \) have global maxima or minima?

Global maxima or minima refer to the highest or lowest values of the function over its entire domain.

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