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I =
0.5 | x3e?x2dx |
0 |
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SHARING.... -experiences of teaching maths as a hobby for over 50 years to school and college students. -challenges in project management and the common sense approach to it from old classics.
0.5 | x3e?x2dx |
0 |
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 |
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 |