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