math
posted by Frank .
Open boxes are being designed to hold mail for the post office. The boxes will be made to have the greatest possible volume. Each box will be made from a sheet of metal that measures 15 inches by 12 inches. They will be formed by cutting and removing a square from each corner. The sides will then be folded up and welded together along each corner. What is the maximum volume?

Let each side of the square to be cut out be x inches
length of base = 152x
width of base = 122x
height of box = x
where 0 < x < 6 or else the dimensions make no sense
Volume = x(152x)(122x)
= 180x  54x^2 + 4x^3
d(Volume)/dx = 180  108x + 12x^2
= 0 for a max/min of Volume
12x^2  108x + 180 = 0
x^2  9x + 15 = 0
x = (9 ± √21)/2
x = 2.2087 or x = 20.62 , but x < 6
so x = 2.2087
Maximum Volume = 2.2087(10.5826)(7.5826) = 177.2340895 cubic inches
check:
let x = 2.2
Volume = 2.2(10.6)(7.6) = 177.232 , a bit smaller
let x = 2.21
Volume = 2.21(10.58)(7.58) = 177.234044 a tiny bit smaller