Posted by **Frank** on Tuesday, February 14, 2012 at 5:33pm.

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?

- math -
**Reiny**, Tuesday, February 14, 2012 at 7:18pm
Let each side of the square to be cut out be x inches

length of base = 15-2x

width of base = 12-2x

height of box = x

where 0 < x < 6 or else the dimensions make no sense

Volume = x(15-2x)(12-2x)

= 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

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