A box with a square base and an open top must have a volume of 32,000 cm3. Find the dimensions of the box that minimize the amount of material used.

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Let the width of the square base be x and height of the box be y, then

Volume, V is given by,

`V = x^2y`

The area (or amount of material), A is given by,

`A = x^2+4(xy)`

Now we have to find values for x and y, such that it A has a minimum with volume of 32000`cm^3`

`32000 = x^2y`

`y = 32000/x^2`

If we substitute this in A,

`A = x^2+4x(32000/x^2)`

`A =...

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