# If a box with a square base and open top is to have a volume of 4 ft^3,find the dimensions that require the least material. Disregard the thickness of the material and waste in construction.

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If the length of a side of the bottom is x and the height is h, then the volume of the box is given by,

`V = x^2h`

4 = x^2h

Therefore, `h = 4/x^2`

The area of the material needed is A,

`A = x^2+4xh`

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

`A = x^2+16/x`

To find the extreme points (maxima, minima or inflection points) we have to check for the points where the first derivative is zero.

`(dA)/(dx) = 2x-16/x^2`

At extreme points, (dA)/(dx) = 0

`2x-16/x^2 = 0`

`2x^3 -16=0`

`x^3 = 8`

`x = 2`

Therefore at x=2, we have an extreme point to determine whether this is a minimum point we have to evaluate the sign of the second derivative.

`(d^2A)/(dx^2) = 2+32/x^3`

This is positive for every x >0, then we can say that at x=2 , A has a minimum value.

The minimum area, `A_min= 2^2+16/2 = 12`

The minimum area is 12.

The height corresponding to minimum area, h is,

`h = 4/2^2 = 1`

**The dimensions for minimum amount of materials are width of 2 and height of **1, **x=2 and h =1.**