MAXIMA-MINIMA......A DIFFERENT PROBLEM
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For details see post on 26-Feb-2015 at 6:00 am.

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SURFACE AREA = S = 2[XY+YZ+ZX] = 7000
XY+YZ+ZX=3500……………………………….1
EDGE LENGTH = P= 4[X+Y+Z]=440
X+Y+Z=110……………………………..2
VOLUME=V=XYZ.................................3

AT THE OUT SET IT IS CLEAR THAT USING CRITERIA FOR LOCAL MAXIMA/MINMA
THE SOLUTION IF ANY  HAS TO BE AT ...X=Y=Z...BUT THIS IS CLEARLY NOT POSSIBLE SINCE ..THIS LEADS TO .....3X^2=3500....OR...X=34.16 FROM EQN. 1 .AND.......
.....3X=110...OR...X=36.7 FROM EQN.2 WHICH ARE CONTRADICTORY ..
SO THE MAXIMA/MINIMA WILL ONLY BE AT BOUNDARIES WHICH HAVE TO BE EVALUATED...

FROM EQN. 2 ….X = 110-Y-Z………….4
PUTTING EQN. 1
Z= [3500-XY]/[X+Y]………………..5
PUTTING EQN. 5 IN EQN.4
X=110-Y-	[3500-XY]/[X+Y]
Y^2+Y[X-110]+[X^2-110X+3500] =0..
	Y=	0.5*[(110-X)+(-3X^2+220X-1900)^0.5]				…OR….
Y=	0.5*[(110-X)-(-3X^2+220X-1900)^0.5]
WE FIND DISCRIMINANT			D = -3X^2+220X-1900			-
D = -[X-10][3X-190 ] ..=0 ..............AT …X = 10 AND ….X= 190/3=63.3333
SO WE GET THE DIMENSIONS &VOLUME AT THE BOUNDARIES AS FOLLOWS
NOTE THAT THE FUNCTIONS BEING SYMMETRIC IN X,Y,Z , WE CAN TAKE
POINTS AS [10,50,50];[50,50,10];[50,10,50]…AND…
[190/3,70/3,70/3] ; [70/3,190/3,70/3] ; [70/3,70/3,190/3]
X	Y	Z1	Z2	V=
10	50	50	50	25000	MINIMUM
50	50	10	10	25000	MINIMUM
50	10	50	50	25000	MINIMUM
63.33333	23.33333	23.33333	23.33333	34481.481	MAXIMUM
23.33333	63.33333	23.33333	23.33333	34481.481	MAXIMUM
23.33333	23.33333	63.33333	63.33333	34481.481	MAXIMUM