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# Calculus/Critical points - multivariables

Question
Hi Paul,

For Analysis I have these exercises where I need to determine critical points and check if they're minima, maxima or none of them both.

I have f(x,y,z)=x²+y²+z²

I have determined the hessian matrix

gives me a diagonal matrix of 2 2 2

so now I need to get my eigen value and see if it's positive or negative.

I'll take & for the symbol of eigenvalue.

so I have (2-&)^3=0

8-8&-4&²+&^3=0

I can't use the quadratic equation, so I have to rewrite the equation. I thought of using a long division and reduce it into two tersm, however I have a rest term of -3...
What would be the easiest way to determine the value of & here?

Questioner:Bob
Country:Limburg, Belgium
Category:Calculus
Private:No
Subject:Checking which critical points are min/max
Question:

Hi Paul,

For Analysis I have these exercises where I need to determine critical points and check if they're minima, maxima or none of them both.

I have f(x,y,z)=x²+y²+z²

I have determined the hessian matrix

gives me a diagonal matrix of 2 2 2

so now I need to get my eigen value and see if it's positive or negative.

I'll take & for the symbol of eigenvalue.

so I have (2-&)^3=0

8-8&-4&²+&^3=0

I can't use the quadratic equation, so I have to rewrite the equation. I thought of using a long division and reduce it into two tersm, however I have a rest term of -3...
What would be the easiest way to determine the value of & here?
...............................................
I think you are working too hard.  The solution(s) to:

(2 - &)^3 = 0  is & = 2, isn't it?

HOWEVER:

fx = 2x
fy = 2y
fz = 2z

Now you want critical points, points where the gradient is zero:

Grad f = <x, y, z>   [take out the factor of 2.]

and you want  x = 0,  y = 0, z = 0.

From Wikipedia:
Critical points and discriminant

If the gradient of f (i.e. its derivative in the vector sense) is zero at some point x, then f has a critical point (or stationary point) at x. The determinant of the Hessian at x is then called the discriminant. If this determinant is zero then x is called a degenerate critical point of f, this is also called a non-Morse critical point of f. Otherwise it is non-degenerate, this is called a Morse critical point of f.

Second derivative test

Main article: Second partial derivative test

The following test can be applied at a non-degenerate critical point x. If the Hessian is positive definite at x, then f attains a local minimum at x. If the Hessian is negative definite at x, then f attains a local maximum at x. If the Hessian has both positive and negative eigenvalues then x is a saddle point for f (this is true even if x is degenerate). Otherwise the test is inconclusive.

.........................................
[ 2 0 0]
[ 0 2 0]
[ 0 0 2]

and its determinant is 8. (so it is not degenerate, hooray)

Now is this 'positive-definite'?

If we compute  x(T) M X

[x y z]  [ 2 0 0] [x]
[ 0 2 0] [y]
[ 0 0 2] [z]

=  2x^2 + 2y^2 + 2z^2  > 0 for all non-zero <x,y,z>.

So you have a local minimum.

Calculus

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#### Paul Klarreich

##### Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

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I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

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