Question 1162529
If the cardboard measures   by    
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No dimensions.
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A general solution, with dim's x by y and the squares removed are z by z in any units.
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Volume = z*(x-2z)*(y-2z) --- x & y are constants, the given dim's
Vol = z*xy - 2z^2*(x+y) + 4z^3
dV/dz = xy - 4z*(x+y) + 12z^2  ---- 1st derivative of volume wrt z
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12z^2 - 4z*(x+y) + xy = 0
z = 4(x+y)/24 + sqrt(16(x+y)^2 - 48xy)/24
{{{z = (x+y)/6 + sqrt((x+y)^2 - 3xy)/6}}}
{{{z = (x+y)/6 + sqrt(x^2 - xy + y^2)/6}}}