My Math Forum

My Math Forum (http://mymathforum.com/math-forums.php)
-   Linear Algebra (http://mymathforum.com/linear-algebra/)
-   -   Fastest structured way to get max(abc) if a+b+c=30 (http://mymathforum.com/linear-algebra/345651-fastest-structured-way-get-max-abc-if-b-c-30-a.html)

kelsiu January 21st, 2019 06:26 PM

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.

skipjack January 22nd, 2019 01:02 AM

Choose a = b = c.

idontknow January 22nd, 2019 05:57 AM

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 .
$\displaystyle (a+b+c)^3 \geq 3^3 abc \;\;$ , $\displaystyle abc\leq 10^3 =max(abc)$

v8archie January 22nd, 2019 06:08 AM

Find the minimum in the reals and test the nearest natural solutions. The minimum product should be fairly obvious.

idontknow January 22nd, 2019 12:39 PM

Other expressions are: $\displaystyle max(abc)=\lfloor (\frac{n}{3})^3 \rfloor $ .

Denis January 22nd, 2019 02:02 PM

10 + 10 + 10 = 30
10 * 10 * 10 = 1000

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

Get my drift?

romsek January 23rd, 2019 10:18 PM

Quote:

Originally Posted by kelsiu (Post 604726)
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)

$3a = 30$
$a=b=c = 10$

skipjack January 24th, 2019 01:25 AM

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.

zylo January 24th, 2019 07:07 AM

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.

idontknow January 24th, 2019 07:24 AM

How to find the minimal value (a fast way)?


All times are GMT -8. The time now is 11:02 AM.

Copyright © 2019 My Math Forum. All rights reserved.