|
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?
|
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)
|
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.
|
Just a heads up as I understand it HW help threads aren't allowed here.
|
|
|
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.
|
Osaka27158 Posts
|
United States24753 Posts
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.
|
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.
|
Aotearoa39261 Posts
(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!
|
17046 Posts
By the way, TL isn't a homework help forum :<
|
Aotearoa39261 Posts
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
|
United States4053 Posts
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
|
Aotearoa39261 Posts
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
|
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.
|
Aotearoa39261 Posts
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
|
On behalf of the 99%, I'd just like to say: Thanks for making us feel stupid, guys.
|
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.
|
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.
|
Aotearoa39261 Posts
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.
|
|
|
|
|
|
EPT
