SOLUTION: (20x)/(102-x) for 0<=x<=100
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Question 51268
:
(20x)/(102-x) for 0<=x<=100
Answer by
THANApHD(104)
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let
20x/102-x <= y
20x <= 102y-xy
20x+xy <= 102y
X(20+y) <= 102y
x <= 102y/20+y
but x <=100
so its obvious 100 = 102y/20+y
2000+100y =102y
2y =2000
y=1000
and other case the least value of the 20x/102-x is 0
so
0<=20x/102-x<=1000