What are the extrema and saddle points of #f(x,y) = xy(e^(y^2)-e^(x^2))#?

Answer 1

# {: ("Critical Point","Conclusion"), ((0,0,0),"saddle") :} #

The theory to identify the extrema of #z=f(x,y)# is:

  1. Solve simultaneously the critical equations

    # (partial f) / (partial x) =(partial f) / (partial y) =0 \ \ \ # (ie #f_x=f_y=0#)

  2. Evaluate #f_(x x), f_(yy) and f_(xy) (=f_(yx))# at each of these critical points. Hence evaluate #Delta=f_(x x)f_(yy)-f_(xy)^2# at each of these points
  3. Determine the nature of the extrema;

    #{: (Delta>0, "There is minimum if " f_(x x)<0),(, "and a maximum if " f_(yy)>0), (Delta<0, "there is a saddle point"), (Delta=0, "Further analysis is necessary") :}#

So we have:

# f(x,y) = xy(e^(y^2)-e^(x^2)) #
# " " = xye^(y^2) - xye^(x^2) #

Let us find the first partial derivatives:

# (partial f) / (partial x) = ye^(y^2) + {(-xy)(2xe^(x^2)) + (-y)(e^(x^2))}#
# \ \ \ \ \ \ \= ye^(y^2) -2x^2ye^(x^2) -ye^(x^2)#

# (partial f) / (partial y) = {(xy)(2ye^(y^2)) + (x)(e^(y^2))} - xe^(x^2)#
# \ \ \ \ \ \ \= 2xy^2e^(y^2) + xe^(y^2) - xe^(x^2)#

So our critical equations are:

# ye^(y^2) -2x^2ye^(x^2) -ye^(x^2) = 0 => y(e^(y^2) -2x^2e^(x^2) -e^(x^2)) = 0#
# 2xy^2e^(y^2) + xe^(y^2) - xe^(x^2) = 0 => x(2y^2e^(y^2) + e^(y^2) - e^(x^2)) = 0#

From the these equations we have:

# y=0# or #e^(y^2) -e^(x^2) = 2x^2e^(x^2)#
# x=0# or #e^(y^2) - e^(x^2) = -2y^2e^(y^2) #

And the only simultaneous solution is #x=y=0#

And so we have one critical point at the origin

So, now let us look at the second partial derivatives so that we can determine the nature of the critical point (I'll just quote these results):

# \ \ \ (partial^2f) / (partial x^2) = -4x^3ye^(x^2)-6xye^(x^2)#
# \ \ \ (partial^2f) / (partial y^2) = 4xy^3e^(y^2)+6xye^(y^2)#
# (partial^2f) / (partial x partial y) = e^(y^2)-e^(x^2)-2x^2e^(x^2)+2y^2e^(y^2) \ \ \ \ (=(partial^2f) / (partial y partial x))#

And we must calculate:

#Delta=(partial^2f) / (partial x^2) (partial^2f) / (partial y^2) - ((partial^2f) / (partial x partial y))^2 #

at each critical point. The second partial derivative values, #Delta#, and conclusion are as follows:

# {: ("Critical Point",(partial^2f) / (partial x^2),(partial^2f) / (partial y^2),(partial^2f) / (partial x partial y),Delta,"Conclusion"), ((0,0,0),0,0,0,= 0,"incluclusive") :} #

So after all that work it is rather disappointing to get an inclusive result, but if we examine the behaviour around the critical point we can readily establish that it is a saddle point.

We can see these critical points if we look at a 3D plot:

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Answer 2

To find the extrema and saddle points of ( f(x,y) = xy(e^{y^2} - e^{x^2}) ), we need to find the critical points where the gradient is zero and classify them using the second partial derivative test.

First, compute the partial derivatives: [ \frac{\partial f}{\partial x} = y(e^{y^2} - 2xe^{x^2}) ] [ \frac{\partial f}{\partial y} = x(2ye^{y^2} - e^{x^2}) ]

Set these partial derivatives to zero to find the critical points.

For ( \frac{\partial f}{\partial x} = 0 ): [ y(e^{y^2} - 2xe^{x^2}) = 0 ] This gives two cases:

  1. ( y = 0 )
  2. ( e^{y^2} = 2xe^{x^2} )

For ( \frac{\partial f}{\partial y} = 0 ): [ x(2ye^{y^2} - e^{x^2}) = 0 ] This gives two cases:

  1. ( x = 0 )
  2. ( 2ye^{y^2} = e^{x^2} )

Now, classify the critical points using the second partial derivative test: [ \frac{\partial^2 f}{\partial x^2} = -2y^2xe^{x^2} ] [ \frac{\partial^2 f}{\partial y^2} = 2x^2e^{y^2} + 2e^{y^2} - 4x^2y^2e^{x^2} ] [ \frac{\partial^2 f}{\partial x \partial y} = (e^{y^2} - 2xe^{x^2}) + y(2ye^{x^2} - 2xe^{y^2}) ]

Evaluate these second partial derivatives at each critical point to determine the nature of the points:

  1. Critical point: ( (0,0) )

    • ( \frac{\partial^2 f}{\partial x^2}(0,0) = 0 )
    • ( \frac{\partial^2 f}{\partial y^2}(0,0) = 2 )
    • ( \frac{\partial^2 f}{\partial x \partial y}(0,0) = 0 )

    The test is inconclusive for this point.

  2. Critical point: ( (x_0,0) ) (where ( e^{x_0^2} = 2x_0 ))

    • ( \frac{\partial^2 f}{\partial x^2}(x_0,0) < 0 )
    • ( \frac{\partial^2 f}{\partial y^2}(x_0,0) > 0 )
    • ( \frac{\partial^2 f}{\partial x \partial y}(x_0,0) = 2x_0 )

    This is a saddle point.

  3. Critical point: ( (0,y_0) ) (where ( 2y_0e^{y_0^2} = 1 ))

    • ( \frac{\partial^2 f}{\partial x^2}(0,y_0) < 0 )
    • ( \frac{\partial^2 f}{\partial y^2}(0,y_0) > 0 )
    • ( \frac{\partial^2 f}{\partial x \partial y}(0,y_0) = 1 )

    This is a saddle point.

In summary:

  • There are two saddle points at ( (x_0,0) ) and ( (0,y_0) ), where ( e^{x_0^2} = 2x_0 ) and ( 2y_0e^{y_0^2} = 1 ), respectively.
  • The critical point at ( (0,0) ) is inconclusive and could be a saddle point or a maximum/minimum, requiring further investigation.
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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