August 26th, 2018, 01:31 AM  #1 
Member Joined: Oct 2012 Posts: 64 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, 08:39 AM  #2  
Math Team Joined: Jul 2011 From: Texas Posts: 2,767 Thanks: 1422  Quote:
If this is not the case, then ignore this solution. $a+b=1 \implies b = 1a$ $y = a^{1a}+(1a)^a + a^a \cdot (1a)^{1a}$ $\log{y} = (1a)\log{a} + a\log(1a) + a\log{a} + (1a)\log(1a)$ $\log{y} = \log{a} + \log(1a)$ $\log{y} = \log[a(1a)]$ $y = a(1a)$ $\dfrac{dy}{da} = 12a = 0 \implies a = \dfrac{1}{2} = b$ $\dfrac{d^2y}{da^2} = 2 \implies y$ is a maximum at $a = \dfrac{1}{2} = b$ $y_{max} = \sqrt{2}+\dfrac{1}{2}$  
August 26th, 2018, 09:27 AM  #3  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,888 Thanks: 765 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan Last edited by skipjack; August 26th, 2018 at 07:16 PM.  
August 26th, 2018, 10:19 AM  #4 
Senior Member Joined: Sep 2015 From: USA Posts: 2,100 Thanks: 1093 
I'm seeing that $\lim \limits_{a\to 0} a^{1a}+(1a)^a + a^a(1a)^{1a} = 2$ and $\lim \limits_{a\to 1} a^{1a}+(1a)^a + a^a(1a)^{1a} = 2$ 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, 10:31 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 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: $\displaystyle y_{\max}=y\left(\frac{1}{2},\frac{1}{2}\right)= \sqrt{2}+\frac{1}{2}$ 
August 26th, 2018, 12:00 PM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 2,767 Thanks: 1422  yep ... I screwed it up. Going back to bed.
Last edited by skipjack; August 26th, 2018 at 07:17 PM. 

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