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[Math Help] AMGM Inequality for Optimization

Blogs > OnceKing
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1 2 Next All
OnceKing
Profile Blog Joined January 2009
United States939 Posts
November 17 2011 05:41 GMT
#1
For my calculus homework today, I was asked to find the value of a such that the curvature of y = e^(ax) was greatest for x = 0.
So I solved this problem probably the 'normal' way -- plug into the curvature function, set x = 0, use the first derivative to maximize the value of the function and find a's value.

It turned out to be a very ugly function to be taking the derivative of (in my opinion, I suppose), and I was thinking of trying to use the Arithmetic Mean-Geometric Mean Inequality to find the maximum value instead. However, I couldn't do it and clearly I can conclude that I really don't understand it at all. So, the question I pose to you is: how would I use the AMGM Inequality to optimize this? Better yet, if someone can give me a complete explanation for how to use it for optimization in general that would be very awesome and balleriffic.

I understand the basic idea: AM >= GM, so for the maximum value we would set AM = GM and solve. But with this problem I was just completely confused with how to do it.

Shown below is what the curvature function ended up being with x = 0.
K(0) = a^2/(1+a^2)^(3/2)

What I tried to do
+ Show Spoiler +
I'm solving for a in a^2/(1+a^2)^(3/2), then I can use the AMGM inequality.
So, if I let a_1 = a^2 and a_2 = (1+a^2)^(-3/2), then I know (a_1+a_2)/2 >= sqrt(a_1*a_2).
And then solving for a becomes a huge mess.
Is it possible that in this case, AMGM is just an undesirable and inefficient way compared to the derivative method for optimization?


"Every man has his follies - and often they are the most interesting thing he has got."
xxpack09
Profile Blog Joined September 2010
United States2160 Posts
November 17 2011 05:49 GMT
#2
I don't even understand how AM/GM is supposed to apply to this at all, I would just use the standard optimization techniques.

btw expect this thread to be closed soon (homework)
Steel
Profile Blog Joined April 2010
Japan2283 Posts
November 17 2011 06:01 GMT
#3
I agree, just use the derivative and practice your algebra Derivatives are always relatively simple to do (integrals are the ugly unsolvable ones )...just need more practice I think.
Try another route paperboy.
Navillus
Profile Joined February 2011
United States1188 Posts
November 17 2011 06:03 GMT
#4
Just a heads up as I understand it HW help threads aren't allowed here.
"TL gives excellent advice 99% of the time. The problem is no one listens to it." -Plexa
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2011-11-17 06:11:16
November 17 2011 06:06 GMT
#5
To bring some closure to this topic, I tried all the AM-GM I know and it doesn't work. For optimization, Jensen's Inequality and Cauchy-Schwarz are usually more well-suited, although I don't think either of those are particularly effective here either.

EDIT: Ok if mani is cool with it, I guess this is fine
Translator:3
OnceKing
Profile Blog Joined January 2009
United States939 Posts
November 17 2011 06:06 GMT
#6
Okay I should quantify this thread then.

This is NOT a homework thread. I am not asking you or anyone to do my homework for me. It involves a homework problem, yes, but I solved it as stated in the OP using the derivative method and got the right answer (sqrt 2, -sqrt 2). This is instead an exploration into solving a problem in an alternative method.
"Every man has his follies - and often they are the most interesting thing he has got."
Manifesto7
Profile Blog Joined November 2002
Osaka27150 Posts
November 17 2011 06:07 GMT
#7
Im cool with it.
ModeratorGodfather
micronesia
Profile Blog Joined July 2006
United States24698 Posts
November 17 2011 06:11 GMT
#8
Yea I left it alone also lol :p

The OP put isn't asking us to do anything... it's more of a math Q&A initially caused by a hw problem than a "do my hw" thread.
ModeratorThere are animal crackers for people and there are people crackers for animals.
igotmyown
Profile Blog Joined April 2009
United States4291 Posts
Last Edited: 2011-11-17 06:45:13
November 17 2011 06:17 GMT
#9
Local maximum and minimum occur when the derivative = 0.
Derivative of a^2/(1+a^2)^(3/2) is
2a/(1+a^2)^(5/2)

Edit: oops
(2a+2a^3-3a^3)/(1+a^2)^(5/2)
2a-a^3=0
a=0 or a=sqrt(1/2), a=-sqrt(1/2)

that's 0 when a = 0. Check whether that's minimum or maximum

I don't remember if that's really the curvature function, so I'm assuming your K(0) is correct.
Plexa
Profile Blog Joined October 2005
Aotearoa39261 Posts
Last Edited: 2011-11-17 06:40:55
November 17 2011 06:19 GMT
#10
(a_1+a_2)/2 >= sqrt(a_1*a_2), max when equal
a_1^2+2a_1*a_2+a_2^2 = 4*a_1*a_2
a_1^2-2a_1*a_2+a_2^2 = 0

Then :
(a_1-a_2)^2=0 => a_1 = a_2 (as expected)

Now: a^2 = (1+a^2)^(-3/2) which apparently doesn't give the right answer!
Administrator~ Spirit will set you free ~
Empyrean
Profile Blog Joined September 2004
16992 Posts
November 17 2011 06:21 GMT
#11
By the way, TL isn't a homework help forum :<
Moderator
Plexa
Profile Blog Joined October 2005
Aotearoa39261 Posts
November 17 2011 06:22 GMT
#12
On November 17 2011 15:11 micronesia wrote:
Yea I left it alone also lol :p

The OP put isn't asking us to do anything... it's more of a math Q&A initially caused by a hw problem than a "do my hw" thread.

It's cool empyrean
Administrator~ Spirit will set you free ~
infinitestory
Profile Blog Joined April 2010
United States4053 Posts
Last Edited: 2011-11-17 06:27:05
November 17 2011 06:24 GMT
#13
ok so I actually found a way to do this, but it's pretty convoluted

Considering the partial fraction decomposition of a^2/(1+a^2)^2 = 1/(1+a^2)-1/(1+a^2)^2 led me to a^2/(1+a^2)^(3/2) = 1/(1+a^2)^(1/2) - 1/(1+a^2)^(3/2). Let x=1/(1+a^2)^(1/2), and so we now want to maximize x-x^3, which is equal to x(1-x^2) or 2*((1/2-(1/2)*x^2)*(x^2)^(1/2). By the weighted form of AM-GM, (((1/2-(1/2)*x^2)*(x^2)^(1/2))^(2/3) <= (2/3)((1/2-(1/2)*x^2)+(1/2)*x^2) = 1/3. This maximum can only be achieved when (1/2-(1/2)*x^2)=(x^2), or x^2=1/3. Substituting gives 1/(1+a^2) = 1/3, or a^2 = 2.

On November 17 2011 15:19 Plexa wrote:
Now: a^2 = (1+a^2)^(-3/2) which you can solve as
0=a^2-(1+a^2)^(2/3) is a difference of two squares

according to Mathematica this gives the wrong answer
Translator:3
Plexa
Profile Blog Joined October 2005
Aotearoa39261 Posts
November 17 2011 06:30 GMT
#14
On November 17 2011 15:24 infinitestory wrote:
ok so I actually found a way to do this, but it's pretty convoluted

Considering the partial fraction decomposition of a^2/(1+a^2)^2 = 1/(1+a^2)-1/(1+a^2)^2 led me to a^2/(1+a^2)^(3/2) = 1/(1+a^2)^(1/2) - 1/(1+a^2)^(3/2). Let x=1/(1+a^2)^(1/2), and so we now want to maximize x-x^3, which is equal to x(1-x^2) or 2*((1/2-(1/2)*x^2)*(x^2)^(1/2). By the weighted form of AM-GM, (((1/2-(1/2)*x^2)*(x^2)^(1/2))^(2/3) <= (2/3)((1/2-(1/2)*x^2)+(1/2)*x^2) = 1/3. This maximum can only be achieved when (1/2-(1/2)*x^2)=(x^2), or x^2=1/3. Substituting gives 1/(1+a^2) = 1/3, or a^2 = 2.

Show nested quote +
On November 17 2011 15:19 Plexa wrote:
Now: a^2 = (1+a^2)^(-3/2) which you can solve as
0=a^2-(1+a^2)^(2/3) is a difference of two squares

according to Mathematica this gives the wrong answer

Something is going funny, lemme check over things
Administrator~ Spirit will set you free ~
igotmyown
Profile Blog Joined April 2009
United States4291 Posts
Last Edited: 2011-11-17 06:40:21
November 17 2011 06:39 GMT
#15
On November 17 2011 15:19 Plexa wrote:
(a_1+a_2)/2 >= sqrt(a_1*a_2), max when equal
a_1^2+2a_1*a_2+a_2^2 = 4*a_1*a_2
a_1^2-2a_1*a_2+a_2^2 = 0

Then :
(a_1-a_2)^2=0 => a_1 = a_2

Now: a^2 = (1+a^2)^(-3/2) which apparently doesn't give the right answer!

So you proved that the geometric mean is maximized relative to the arithmetic mean when it's equal.

Then you assumed the numerator and denominator of a problem are equal, to achieve said maximization.
Plexa
Profile Blog Joined October 2005
Aotearoa39261 Posts
November 17 2011 06:46 GMT
#16
On November 17 2011 15:39 igotmyown wrote:
Show nested quote +
On November 17 2011 15:19 Plexa wrote:
(a_1+a_2)/2 >= sqrt(a_1*a_2), max when equal
a_1^2+2a_1*a_2+a_2^2 = 4*a_1*a_2
a_1^2-2a_1*a_2+a_2^2 = 0

Then :
(a_1-a_2)^2=0 => a_1 = a_2

Now: a^2 = (1+a^2)^(-3/2) which apparently doesn't give the right answer!

So you proved that the geometric mean is maximized relative to the arithmetic mean when it's equal.

Then you assumed the numerator and denominator of a problem are equal, to achieve said maximization.

It doesn't take very long to realise that there is something amiss, but I think the bigger question is understanding why you wouldn't go for this approach as opposed to infinitestory's
Administrator~ Spirit will set you free ~
SonuvBob
Profile Blog Joined October 2006
Aiur21549 Posts
November 17 2011 06:47 GMT
#17
On behalf of the 99%, I'd just like to say: Thanks for making us feel stupid, guys.
Administrator
igotmyown
Profile Blog Joined April 2009
United States4291 Posts
Last Edited: 2011-11-17 06:49:09
November 17 2011 06:48 GMT
#18
On November 17 2011 15:24 infinitestory wrote:
ok so I actually found a way to do this, but it's pretty convoluted

Considering the partial fraction decomposition of a^2/(1+a^2)^2 = 1/(1+a^2)-1/(1+a^2)^2 led me to a^2/(1+a^2)^(3/2) = 1/(1+a^2)^(1/2) - 1/(1+a^2)^(3/2). Let x=1/(1+a^2)^(1/2), and so we now want to maximize x-x^3, which is equal to x(1-x^2) or 2*((1/2-(1/2)*x^2)*(x^2)^(1/2). By the weighted form of AM-GM, (((1/2-(1/2)*x^2)*(x^2)^(1/2))^(2/3) <= (2/3)((1/2-(1/2)*x^2)+(1/2)*x^2) = 1/3. This maximum can only be achieved when (1/2-(1/2)*x^2)=(x^2), or x^2=1/3. Substituting gives 1/(1+a^2) = 1/3, or a^2 = 2.

Show nested quote +
On November 17 2011 15:19 Plexa wrote:
Now: a^2 = (1+a^2)^(-3/2) which you can solve as
0=a^2-(1+a^2)^(2/3) is a difference of two squares

according to Mathematica this gives the wrong answer


A very convoluted way to prove you maximize x-x^3 at -1 and 1, and you also aren't allowed to let x<0 or 1-x^2 <0 for the inequality to hold. But clever.
igotmyown
Profile Blog Joined April 2009
United States4291 Posts
November 17 2011 06:54 GMT
#19
On November 17 2011 15:46 Plexa wrote:
Show nested quote +
On November 17 2011 15:39 igotmyown wrote:
On November 17 2011 15:19 Plexa wrote:
(a_1+a_2)/2 >= sqrt(a_1*a_2), max when equal
a_1^2+2a_1*a_2+a_2^2 = 4*a_1*a_2
a_1^2-2a_1*a_2+a_2^2 = 0

Then :
(a_1-a_2)^2=0 => a_1 = a_2

Now: a^2 = (1+a^2)^(-3/2) which apparently doesn't give the right answer!

So you proved that the geometric mean is maximized relative to the arithmetic mean when it's equal.

Then you assumed the numerator and denominator of a problem are equal, to achieve said maximization.

It doesn't take very long to realise that there is something amiss, but I think the bigger question is understanding why you wouldn't go for this approach as opposed to infinitestory's


But what you did was found a way to maximize a^2+(1+a)^2)^(--3/2) (or some variant of the sum), relative to the product, rather than maximize the product itself.
Plexa
Profile Blog Joined October 2005
Aotearoa39261 Posts
November 17 2011 06:57 GMT
#20
On November 17 2011 15:54 igotmyown wrote:
Show nested quote +
On November 17 2011 15:46 Plexa wrote:
On November 17 2011 15:39 igotmyown wrote:
On November 17 2011 15:19 Plexa wrote:
(a_1+a_2)/2 >= sqrt(a_1*a_2), max when equal
a_1^2+2a_1*a_2+a_2^2 = 4*a_1*a_2
a_1^2-2a_1*a_2+a_2^2 = 0

Then :
(a_1-a_2)^2=0 => a_1 = a_2

Now: a^2 = (1+a^2)^(-3/2) which apparently doesn't give the right answer!

So you proved that the geometric mean is maximized relative to the arithmetic mean when it's equal.

Then you assumed the numerator and denominator of a problem are equal, to achieve said maximization.

It doesn't take very long to realise that there is something amiss, but I think the bigger question is understanding why you wouldn't go for this approach as opposed to infinitestory's


But what you did was found a way to maximize a^2+(1+a)^2)^(--3/2) (or some variant of the sum), relative to the product, rather than maximize the product itself.

Ya, I see it.
Administrator~ Spirit will set you free ~
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