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August 26th, 2018, 01:31 AM   #1
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maximum value

let a,b real numbers such a+b=1. what is the max value of
a^b + b^a + (a^a)(b^b)
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August 26th, 2018, 08:39 AM   #2
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Quote:
Originally Posted by fahad nasir View Post
let a,b real numbers such a+b=1. what is the max value of
a^b + b^a + (a^a)(b^b)
Since $a$ and $b$ are bases of exponential terms, I am restricting their values to the positive reals $\implies$ both $a$ and $b$ lie in the open interval $(0,1)$.

If this is not the case, then ignore this solution.

$a+b=1 \implies b = 1-a$

$y = a^{1-a}+(1-a)^a + a^a \cdot (1-a)^{1-a}$

$\log{y} = (1-a)\log{a} + a\log(1-a) + a\log{a} + (1-a)\log(1-a)$

$\log{y} = \log{a} + \log(1-a)$

$\log{y} = \log[a(1-a)]$

$y = a(1-a)$

$\dfrac{dy}{da} = 1-2a = 0 \implies a = \dfrac{1}{2} = b$

$\dfrac{d^2y}{da^2} = -2 \implies y$ is a maximum at $a = \dfrac{1}{2} = b$

$y_{max} = \sqrt{2}+\dfrac{1}{2}$
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August 26th, 2018, 09:27 AM   #3
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Quote:
Originally Posted by skeeter View Post
$y = a^{1-a}+(1-a)^a + a^a \cdot (1-a)^{1-a}$

$\log{y} = (1-a)\log{a} + a\log(1-a) + a\log{a} + (1-a)\log(1-a)$
I haven't taken my meds yet today, but last I knew $\displaystyle \log(a + b) \neq \log(a) + \log(b)$.

-Dan
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Last edited by skipjack; August 26th, 2018 at 07:16 PM.
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August 26th, 2018, 10:19 AM   #4
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I'm seeing that

$\lim \limits_{a\to 0} a^{1-a}+(1-a)^a + a^a(1-a)^{1-a} = 2$

and

$\lim \limits_{a\to 1} a^{1-a}+(1-a)^a + a^a(1-a)^{1-a} = 2$

The reason this doesn't correspond to the maximum found by skeeter is that the domain is restricted thus creating a boundary on which the maxima lie.
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August 26th, 2018, 10:31 AM   #5
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Using the notion of cyclic symmetry, I get the same critical point as skeeter, namely:

$\displaystyle (a,b)=\left(\frac{1}{2},\frac{1}{2}\right)$

And I agree that:

$\displaystyle y\left(\frac{1}{2},\frac{1}{2}\right)= \sqrt{2}+\frac{1}{2}$

Checking the objective function at another point on the constraint, we find:

$\displaystyle y\left(\frac{3}{4},\frac{1}{4}\right)<\sqrt{2}+ \frac{1}{2}$

And so I would also conclude:

$\displaystyle y_{\max}=y\left(\frac{1}{2},\frac{1}{2}\right)= \sqrt{2}+\frac{1}{2}$
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August 26th, 2018, 12:00 PM   #6
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Quote:
Originally Posted by topsquark View Post
I haven't taken my meds yet today, but last I knew $\displaystyle \log(a + b) \neq \log(a) + \log(b)$.

-Dan
yep ... I screwed it up. Going back to bed.
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Last edited by skipjack; August 26th, 2018 at 07:17 PM.
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