see diagram below.

Let the sides of the squares cut out be x

then V = lwh

=> V = (3 - 2x)(3 - 2x)x

=> V = (3 - 2x)^2*x

=> V = (9 - 12x + 4x^2)x

=> V = 4x^3 - 12x^2 + 9x

=> V' = 12x^2 - 24x + 9

for max V, set V' = 0

=> 12x^2 - 24x + 9 = 0

=> 4x^2 - 12x + 3 = 0

=> (2x - 3)(2x - 1) = 0

=> 2x - 3 = 0 or 2x - 1 = 0

=> x = 3/2 or x = 1/2

now V'' = 24x - 24

when x = 3/2

V'' = 24(3/2) - 24 = 12 > 0

this is a concave up, so this is a local min, so this is the minimum value based on the second derivative test

when x = 1/2

V'' = 24(1/2) - 24 = -12 < 0

this is a concave down, so this is a local max, so this is the maximum value based on the second derivative test

so for the max Volume, which we will call maxV, x = 1/2

but V = (3 - 2x)^2 * x

=> maxV = (3 - 2(1/2))^2 * (1/2)

=> maxV = 2 cm^2

EDIT: oh, sorry, we're working in ft, just change all the cm to ft in the diagram and my calculations