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January 21st, 2019, 07:26 PM   #1
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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 07:29 PM.
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January 22nd, 2019, 02:02 AM   #2
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Choose a = b = c.
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January 22nd, 2019, 06:57 AM   #3
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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 07:11 AM.
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January 22nd, 2019, 07:08 AM   #4
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Find the minimum in the reals and test the nearest natural solutions. The minimum product should be fairly obvious.
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January 22nd, 2019, 01:39 PM   #5
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Other expressions are: $\displaystyle max(abc)=\lfloor (\frac{n}{3})^3 \rfloor $ .

Last edited by idontknow; January 22nd, 2019 at 01:45 PM.
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January 22nd, 2019, 03:02 PM   #6
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10 + 10 + 10 = 30
10 * 10 * 10 = 1000

9.99 + 10 + 10.01 = 30
9.99 * 10 * 10.01 = 999.99899....

Get my drift?
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January 23rd, 2019, 11:18 PM   #7
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Quote:
Originally Posted by kelsiu View Post
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)

$3a = 30$
$a=b=c = 10$
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January 24th, 2019, 02:25 AM   #8
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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.
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January 24th, 2019, 08:07 AM   #9
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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 12:42 PM.
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January 24th, 2019, 08:24 AM   #10
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How to find the minimal value (a fast way)?
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