Critical Points and the Second Derivative Test for a Multivariable Function

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The discussion focuses on finding critical points and applying the second derivative test for the function f(x,y)=x^4+y^4+4xy. The user calculates the gradient, sets it to zero, and finds critical points at (0,0) and (1,-1), but overlooks additional roots, including (-1,1). Clarification is sought on whether the Hessian matrix is involved in the second derivative test, and confusion arises regarding the complexity of the determinant calculation. The conversation concludes with an acknowledgment of the missed critical point and unnecessary complications involving eigenvalues.
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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!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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