v = the vertical height of the turned up edge

b = the width of the remaining base.

2*v + b = 64

v * b = A (objective Amax)

Let’s simplify the second equation by substituting b from the 1st:

b = 64 – 2*v, so

v * (64 – 2*v) = A, or

64*v – 2*v^2 = A(v)

We want to know the maximum A(v), which can be found where the derivative of A(v) with respect to v is 0:

d[A(v)] = 64 – 4*v,

where 64 – 4*v = 0, v = 16

Test this: using equation 1, 2*v + b = 64, or 2 * 16 +b = 64, b = 32

So Amax would be v * b, or 16 * 32 = 512 cm^2, (whereas dividing the sheet in 3rds, giving height and width of 21.33 cm will only provide 455 cm^2)

Since the gutter will have 3 equal sides… 64cm/3 = 21.33 cm…

so you should turn up 21.33 cms on each side…