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...

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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 = x^2+128000/x`

Let's find the derivative of A wrt x,

`(dA)/(dx) = 2x - 128000/x^2`

For extreme points (Maximums or minimums or inflection points), first derivative must be zero.

Therefore,

`0 = 2x -128000/x^2`

`x = 64000/x^2`

`x^3 = 64000`

`x^3 -40^3 = 0`

`(x-40)(x^2+40x+40^2) = 0`

Therefore, `x = 40` .

To check whether this is a minimum, you have to find the second derivative.

`(d^2A)/(dx^2) = 2 -(-2*128000/x^3)`

`(d^2A)/(dx^2) = 2 +256000/x^3`

at x = 40,

`(d^2A)/(dx^2) = 2 +256000/40^3` `gt 0`

Therefore, A has a minimum at x =40.

Therefore y is,

`y = 32000/x^2`

`y = 32000/40^2 = 32000/1600 = 20`

**The dimensions of the box so that the amount of material used is minimum are width = 40 cm and height = 20 cm.**