Posted by Kieran on Thursday, September 27, 2012 at 9:24am.
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+-3)
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...
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