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January 24th, 2019, 08:50 AM   #11
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Quote:
Originally Posted by idontknow View Post
How to find the minimal value (a fast way)?
There isn't any. The method, Romsek-Lagrange multipliers, and mine-based on principle behind Lagrange multipliers, picks up all extrema.

EDIT:

Actually, the most basic way to look for extrema is to substitute for z to get:
H(x,y)=xy(30-x-y)
Then Hx=Hy=0 gives x=y=z=10.
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Last edited by zylo; January 24th, 2019 at 09:23 AM.
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January 24th, 2019, 09:23 AM   #12
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Quote:
Originally Posted by zylo View Post
There isn't any. The method, Romsek-Lagrange multipliers, and mine-based on principle behind Lagrange multipliers, picks up all extrema.
you can apply the 2nd derivative test to the solutions found and weed out non-maxima

The harder problem is if the maxima is on the boundary in which case it won't be picked up using Lagrange Multipliers at all.
As far as I know in that case you just have to check the boundaries using brute force.
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January 24th, 2019, 09:29 AM   #13
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There is only one solution.

What boundary? There isn't any.

If you accept 0's, the minimum of xyz is x=0, y=0, z=30

EDIT Again:

With that in mind, recalling v8archie's suggestion, the nearest integers would be x=1, y=1, then z=28.
Looks reasonable but can't prove it.

Last edited by skipjack; January 24th, 2019 at 12:48 PM.
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