Applied Calculus
posted by Jacob on .
If an open box is made from a tin sheet 7 in. square by cutting out identical squares from each corner and bending up the resulting flaps, determine the dimensions of the largest box that can be made. (Round your answers to two decimal places.)
Height:
Length:
Width:

base = 7 2x
height = x
volume = x(72x)^2
= 49x  28x^2 + 4x^3
d(volume)/dx = 49  56x + 12x^2
= 0 for a max volume
12x^2  56x + 49 = 0
x = (56 ± √784)/24
= (56 ± 28)/24 = 3.5 or 7/6 or 1.1666...
but clearly x < 3.5 or we have cut the whole base away.
base is 7  2(7/6) = 14/3 by 14/3
and the height is 7/6
round to your required decimals 
new length = 7  2 h
height = h
volume = (72h)(72h)(h)
v = (49 28 h + 4 h^2)h
so
v = 4 h^3 28 h^2 + 49 h
dv/dh = 0 for max or min
dv/dh = 12 h^2 56 h + 49 = 0
(6h 7)(2 h 7) = 0
h = 7/6 or h = 7/2
if h = 7/2, the box has zero bottom
so the answer is h = 7/6
7  2(7/6) = 77/3 = 14/3 = length and width