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.