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 August 26th, 2018, 02:31 AM #1 Member   Joined: Oct 2012 Posts: 71 Thanks: 0 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)
August 26th, 2018, 09:39 AM   #2
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Quote:
 Originally Posted by fahad nasir 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 $a = \dfrac{1}{2} = b$

$y_{max} = \sqrt{2}+\dfrac{1}{2}$

August 26th, 2018, 10:27 AM   #3
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 Originally Posted by skeeter $y = a^{1-a}+(1-a)^a + a^a \cdot (1-a)^{1-a}$ $\log{y} = (1-a)\log{a} + a\log(1-a) + a\log{a} + (1-a)\log(1-a)$
I haven't taken my meds yet today, but last I knew $\displaystyle \log(a + b) \neq \log(a) + \log(b)$.

-Dan

Last edited by skipjack; August 26th, 2018 at 08:16 PM.

 August 26th, 2018, 11:19 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 2,299 Thanks: 1218 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.
 August 26th, 2018, 11:31 AM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 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: $\displaystyle y_{\max}=y\left(\frac{1}{2},\frac{1}{2}\right)= \sqrt{2}+\frac{1}{2}$ Thanks from fahad nasir
August 26th, 2018, 01:00 PM   #6
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Quote:
 Originally Posted by topsquark 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.

Last edited by skipjack; August 26th, 2018 at 08:17 PM.

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