Calc
posted by Kieran .
a box with an open top is to be made from a rectangular piece of tin by cutting equal squares from the corners and turning up the sides. The piece of tin measures 1mx2m. Find the size of the squares that yields a maximum capacity for the box.
So far i have
V=(12x)(22x)x

so, figure dV/dx
dV/dx = 2(6x^26x+1)
dV/dx = 0 when x = 1/6 (3±√3) = .211 or .789
Now .789 is impossible, since the width is only 1.
so, the cuts are .211m 
how did you go from 0= 1/6 (3+ã3)

dV/dx = 2(6x^26x+1
so, dV/dx = 0 when 6x^26x+1
solve the quadratic to get X = 1/6 (3±√3)
this is calculus; algebra I should be no problem...
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