Critical Points and the Second Derivative Test for a Multivariable Function

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Homework Help Overview

The discussion revolves around finding critical points and applying the second derivative test for the multivariable function f(x,y) = x4 + y4 + 4xy. Participants are exploring the process of determining local maxima, minima, or saddle points.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss taking the gradient and setting it to zero to find critical points. There are attempts to solve for y in terms of x and substitute back to find roots. Questions arise about the completeness of the critical points found and the interpretation of the Hessian in the context of the second derivative test.

Discussion Status

Some participants have provided feedback on the identification of critical points, noting that not all roots may have been found. There is a recognition of confusion regarding the Hessian and its determinant, with suggestions to clarify the evaluation process.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information available and the methods they can use. There is an acknowledgment of potential oversight in identifying critical points and the complexity of the calculations involved.

sheepcountme
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Homework Statement



Find the critical points and use the second derivative test to decide if your critical points are local maxima, local minima, or saddle points.

f(x,y)=x4+y4+4xy


The Attempt at a Solution



so I took the gradient to get: <4x3+4y, 4y3+4x>

I know I need to set this equal to <0,0>..so,


4x3+4y=0 and 4y3+4x=0

but I'm stuck...I tried solving for y in the first one to get

y=-x3 and then plugging this into the next equation to get -x9+x=0

If I solve for x, I believe I get x=0 or x=1 and then plugging these into the first I get the points (0,0) and (1,-1)

Have I done this correctly?

And when we're talking about the second derivative test, is this the Hessian? And if so, I've gotten 144x2y2-12x2\lambda-12y2\lambda+\lambda2-16=0

which seems awfully messy to be able to determine if the point is a maxima, etc.
 
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That seems pretty ok, but what's wrong with (-1,1) as a critical point? I don't think you found all of the roots. And what are those lambdas doing the Hessian? Isn't it just the determinant of the matrix of second derivatives?
 
sheepcountme said:

Homework Statement



Find the critical points and use the second derivative test to decide if your critical points are local maxima, local minima, or saddle points.

f(x,y)=x4+y4+4xy


The Attempt at a Solution



so I took the gradient to get: <4x3+4y, 4y3+4x>

I know I need to set this equal to <0,0>..so,


4x3+4y=0 and 4y3+4x=0

but I'm stuck...I tried solving for y in the first one to get

y=-x3 and then plugging this into the next equation to get -x9+x=0

If I solve for x, I believe I get x=0 or x=1 and then plugging these into the first I get the points (0,0) and (1,-1)
x^9- x= x(x^8- 1)= 0
Yes, x= 0 is a root. But x^8- 1= 0 has two real roots.

Have I done this correctly?

And when we're talking about the second derivative test, is this the Hessian? And if so, I've gotten 144x2y2-12x2\lambda-12y2\lambda+\lambda2-16=0

which seems awfully messy to be able to determine if the point is a maxima, etc.
Look at
\left|\begin{array}{cc}f_{xx} &amp; f_{xy} \\ f_{xy} &amp; f_{yy}\end{array}\right|= \left|\begin{array}{cc}12x^3 &amp; 4 \\ 4 &amp; 12 y^3\end{array}\right|
Put in the values of x and y before evaluating the determinant and it is not at all complicated!
 
Ah! I missed (-1,1), thank you!

And I got eigenvalues unnecessarily mixed up in all this (where the lamdas came from), thanks!
 

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