Question

Anyone good in Maxima and Minima of two and more variables? Then I have a question

Answer 1

Ok thanks, I attached the question

Answer 2

did you find the critical point

Answer 3

i think we need to solve
fx = 0
fy =0
fz = 0

Answer 4

Yeah I know that, but I am unable to prove the maximum or minimum part

Answer 5

what do you have so far

Answer 6

I computed the value of the stationary point and the stationary value is what the proof requires. Now I have to show that the stationary value is an extreme value? Will I post the solution till then?

Answer 7

i know how to determine maximum/minima for f(x,y) using a determinant

Answer 8

i bet you that there is a determinant for f(x,y,z)

Answer 9

If you do what @perl said above, i.e. solve:\[f_x=0\]\[f_y=0\]\[f_z=0\]then you should be able to get 3 expressions and take advantage of symmetry to deduce the result.

Answer 10

what did you get for \(f_x\)?

Answer 11

Check the attachment. I got the stationary value. I need to prove that it is an extreme value

Answer 12

that is correct, and since this came as a result of solving:\[f_x=0\]\[f_y=0\]\[f_z=0\]then it must represent an extreme value

Answer 13

No fx=fy=fz=0 is a necessary condition, not a sufficient one

Answer 14

do you mean you want to prove that it is a maxima rather than, say, a minima? In which case I believe you need to employ the second derivative test.

Answer 15

by extreme value you mean this is an absolute minimum or absolute maximum?

Answer 16

Extreme value means maxima or minima. Every extreme value is a stationary value but every stationary value is not an extreme value.

Answer 17

Yup absolute maxima or minima

Answer 18

An extremum (extrema is plural) is a maximum or minimum. An extremum may be local (a.k.a. a relative extremum; an extremum in a given region which is not the overall maximum or minimum) or global.

Answer 19

@perl please solve the problem further. From here I cannot proceed

Answer 20

aren't we done?

Answer 21

you want to prove there are no other extrema?

Answer 22

No see the question. We have to prove that this value is an extreme value

Answer 23

ok - this is indeed more complicated than it first appeared. I have found a reference that may help you @Princer_Jones : http://www.maths.bris.ac.uk/~maxmr/opt/multvar.pdf

Answer 24

I assume you will need to calculate: \(f_{xx}, f_{xy}, f_{xz}, f_{yx},...,f_{zz}\)

Answer 25

Yeah I know this method asnaseer.:)

Answer 26

I am learning it as we type... :)

Answer 27

So I believe this employs something called the Hessian Matrix

Answer 28

Oh..:P

Answer 29

Yeah obviously and the concept is in linear algebra

Answer 30

So which part are you stuck on?

PS: I study maths as a hobby - so I enjoy learning new techniques whenever I can

Answer 31

Dint try after this, I thought there will be any other easier method for that reason only.:P

Answer 32

Oh I see - I am afraid I won't be able to help you there but maybe some others on here will have a better idea. Let me tag them for you.

@ganeshie8 @Zarkon - do you guys know how to solve this type of problem?

Answer 33

How to give a medal, then i will give you one for helping

Answer 34

Don't worry - I am not on here to collect medals. I just enjoy learning and helping where I can :)

Answer 35

@Kainui - do you know how to solve this type of problem?

Answer 36

A thought - since we have shown that the stationary point is at x=y=z, is it "legal" to then simplify the original expression to just:\[f=3\log(3x^2)-6x^3\]and use the second derivative test on this?

Answer 37

No... it is not valid. x=y=z is not for all values.:)

Answer 38

Using Mathematica, I found that your function has a maximum at
\[
\left\{x\to \frac{1}{\sqrt[3]{3}},y\to
\frac{1}{\sqrt[3]{3}},z\to \frac{1}{\sqrt[3]{3}}\right\}
\]
and f evaluated at this point gives you the value
-2 + ln(3) which is your value.
I will think of an easier way how to show this by hand and let you know

Answer 39

I have solved this the stationary part.. Please check the attachment

Answer 40

a stationary point could be :
1) local min
2) local max
3) saddle point

Answer 41

familiar with second derivative test for multiple variables ?

Answer 42

Yeah..

Answer 43

you don't want to use that since its a pain ?

Answer 44

No thats the last method and its too long, 3 pages of calculations... If there is an easy method, I want to know

Answer 45

@everyone - can we not take advantage of the fact that the function is symmetric in x, y and z?

Answer 46

i see... something using number theory or some other clever tricks ?

Answer 47

@asnaseer it need not be always true that x=y=z gives the extrema if the function is symmetric, right ?

Answer 48

Yes.:)

Answer 49

@ganeshie8 - correct, but we have already shown that the stationary point is at \(x=y=z=\frac{1}{\sqrt[3]{3}}\)

Answer 50

Oh right!

Answer 51

what is needed now is to show that this point is a maxima (I believe). And this is where I am wondering if we can somehow take advantage of the symmetry in the function definition?

Answer 52

if we could show that x=y=z is an extrema,

then we can argue that it has to be maximum since \(\ln(\mathcal{O(x^2})) \lt \mathcal{O(x^3)}\)

Answer 53

there is a first derivative test using directional derivatives if you wanted to try that. I'm sure you could find it online.

Answer 54

I need to break for food now but will come back afterwards to see if anyone has managed to find a clever trick here :)

Answer 55

though the 2nd D test is pretty easy so I'm not sure why you are not using that

Answer 56

Just show that the Hessian is negative definite

Answer 57

It is easy but lengthy. And its the last weapon against this problem.:)

Answer 58

it's probably the best weapon too. Nothing wrong with hitting a problem with a sledgehammer.

Answer 59

Here all the points where the gradient vector is zero
\[
\left\{\{x\to 0,y\to 0,z\to 1\},\{x\to 0,y\to 1,z\to
0\}, \\\left\{x\to 0,y\to \frac{1}{\sqrt[3]{2}},z\to
\frac{1}{\sqrt[3]{2}}\right\},\{x\to 1,y\to 0,z\to
0\},\\\left\{x\to \frac{1}{\sqrt[3]{2}},y\to 0,z\to
\frac{1}{\sqrt[3]{2}}\right\},\left\{x\to
\frac{1}{\sqrt[3]{2}},y\to \frac{1}{\sqrt[3]{2}},z\to
0\right\},\\\left\{x\to \frac{1}{\sqrt[3]{3}},y\to
\frac{1}{\sqrt[3]{3}},z\to
\frac{1}{\sqrt[3]{3}}\right\}\right\}
\]
You have to verify that all are not maxima or minima except the last one

Answer 60

I have not read your attachment about the stationary points. According to my post above there are stationary points other than x=y=z

Answer 61

Read my attachment

Answer 62

You are assuming that all variables cannot be zero. We are given that
\[
(x,y,z)\ne (0,0,0)
\]
so (0,y,z) is allowed.

Answer 63

check it,, I already got the stationary point

Answer 64

@eliassaab - well spotted, I also mis-read \((x,y,z)\ne(0,0,0)\) to mean:\[x\ne0\]OR\[y\ne0\]OR\[z\ne0\]when in fact I believe it means they cannot ALL be equal to zero simultaneously.

Answer 65

So, for example, the first derivative with respect to x gives:\[f_x=\frac{6x(1-x(x^2+y^2+z^2))}{x^2+y^2+z^2}\]which means \(f_x=0\) implies:\[x=0\]OR\[x(x^2+y^2+z^2)=1\]which leads to all the stationary points you found above.

So I guess there is no real shortcut to employ here :(

Answer 66

the problem is more complex. now the stationary point calculation is also wrong i guess

Answer 67

The stationary points calculations is right.

Answer 68

Yeah but how.. explain..

Answer 69

I mean (0,y,z) is also possible.

Answer 70

I did the solutions with Mathematica
Solve[grad[f[x, y, z], {x, y, z}] == {0., 0., 0.}, {x, y, z}, Reals]
and it gave the answers above for example
In[768]:= grad[f[x, y, z], {x, y, z}, {0, 1/2^(1/3), 1/2^(1/3)}]

Out[768]= {0, 0, 0}