January 21st, 2019, 06:26 PM  #1 
Newbie Joined: Jun 2014 From: Hong Kong Posts: 7 Thanks: 0  Fastest structured way to get max(abc) if a+b+c=30
What is the fastest and structured way to get maximum of abc if a+b+c=n, say n=30? a,b,c are nonnegative and can be noninteger.
Last edited by kelsiu; January 21st, 2019 at 06:29 PM. 
January 22nd, 2019, 01:02 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,616 Thanks: 2072 
Choose a = b = c.

January 22nd, 2019, 05:57 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 509 Thanks: 79 
What about for a,b,c natural numbers ? And how to find the minimal value? I searched a bit for LagrangeMultipliers but how to use it for natural numbers? For this type of problem AMGM works . $\displaystyle (a+b+c)^3 \geq 3^3 abc \;\;$ , $\displaystyle abc\leq 10^3 =max(abc)$ Last edited by idontknow; January 22nd, 2019 at 06:11 AM. 
January 22nd, 2019, 06:08 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,649 Thanks: 2630 Math Focus: Mainly analysis and algebra 
Find the minimum in the reals and test the nearest natural solutions. The minimum product should be fairly obvious.

January 22nd, 2019, 12:39 PM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 509 Thanks: 79 
Other expressions are: $\displaystyle max(abc)=\lfloor (\frac{n}{3})^3 \rfloor $ .
Last edited by idontknow; January 22nd, 2019 at 12:45 PM. 
January 22nd, 2019, 02:02 PM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,412 Thanks: 1024 
10 + 10 + 10 = 30 10 * 10 * 10 = 1000 9.99 + 10 + 10.01 = 30 9.99 * 10 * 10.01 = 999.99899.... Get my drift? 
January 23rd, 2019, 10:18 PM  #7  
Senior Member Joined: Sep 2015 From: USA Posts: 2,425 Thanks: 1312  Quote:
$\nabla \left(a b c  \lambda(a + b + c  30)\right)$ $b c  \lambda = 0$ $a c  \lambda = 0$ $a b  \lambda = 0$ $a+b+c = 30$ $\dfrac b a = 1$ $\dfrac c a = 1$ $\dfrac c b = 1$ $a = b = c$ (As Skipjack mentioned) $3a = 30$ $a=b=c = 10$  
January 24th, 2019, 01:25 AM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,616 Thanks: 2072 
If p and q are any two of the variables in the original problem, the solution must maximize pq = ((p + q)²  (p  q)²)/4, where p + q is a constant, so p = q. 
January 24th, 2019, 07:07 AM  #9 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 
F(x,y,z)=xyz, G(x,y,z)=x+y+z=30 dF=Fxdx+Fydy+Fzdz=0, but dx, dy, dz not arbitrary. They are subject to dG=dx+dy+dz=0 > (FxFz)dx=0, (FyFz)dy=0, dx and dy arbitrary.> Fx=Fy > yz=xz > x=y But I could also have let dx and dz be arbitrary. Repeating the above I would get x=z. So x=y=z=10. Last edited by skipjack; January 24th, 2019 at 11:42 AM. 
January 24th, 2019, 07:24 AM  #10 
Senior Member Joined: Dec 2015 From: somewhere Posts: 509 Thanks: 79 
How to find the minimal value (a fast way)?


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c30, fastest, maxabc, structured 
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