My Math Forum Fastest structured way to get max(abc) if a+b+c=30

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 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 non-negative and can be non-integer. 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. Thanks from idontknow
 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 Lagrange-Multipliers but how to use it for natural numbers? For this type of problem AM-GM 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. Thanks from idontknow
 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
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Quote:
 Originally Posted by kelsiu What is the fastest and structured way to get maximum of abc if a+b+c=n, say n=30? a,b,c are non-negative and can be non-integer.
Solve

$\nabla \left(a b c - \lambda(a + b + c - 30)\right)$

$b c - \lambda = 0$
$a c - \lambda = 0$
$a b - \lambda = 0$
$a+b+c = 30$

$\dfrac b a = 1$

$\dfrac c a = 1$

$\dfrac c b = 1$

$a = b = c$ (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. Thanks from greg1313
 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 -> (Fx-Fz)dx=0, (Fy-Fz)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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