January 24th, 2019, 07:50 AM  #11 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  There isn't any. The method, RomsekLagrange multipliers, and minebased 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(30xy) Then Hx=Hy=0 gives x=y=z=10. Last edited by zylo; January 24th, 2019 at 08:23 AM. 
January 24th, 2019, 08:23 AM  #12  
Senior Member Joined: Sep 2015 From: USA Posts: 2,588 Thanks: 1430  Quote:
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.  
January 24th, 2019, 08:29 AM  #13 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
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 11:48 AM. 

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c30, fastest, maxabc, structured 
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