max and min problem (minimizing surface area of box)

[Guest]

A rectangular box with a square base and no top is to have volume of 108 cubic inches. Find the dimensions for the box that require the least amount of material.

The function that is to be minimized is the surface area (s) while the volume (v) remains fixed at 108 cubic inches.

Letting x = length of the squre base and h = height of hte box you find that

V= x^2h = 108 cu in
h = 108/x^2

s= x^2 +4xh
s = f(x) = x^2 + 4x(108/x^2)
f(x) = x^2 + 432/x

with the domain of f(x) = (0, +infinity) because x represents a length.
f'(x) = 2x - 432/x^2
f'(x) = 0
2x- 432/x^2 = 0

2x^3 - 432 = 0
2x^3 = 432
x^3 = 216
x = 6

hence, a critical point occurs when x = 6. Using the 2nd derivative test:
f''(x) = 2 + 864/x^3
f''(6) = 6 > 0
and f has a local minimum at x = 6; hence, the dimensions of the box that require the least amount of material are a length and width of 6 inches and a height of 3 inches.

Okay, sorry for the long drawn out problem. But here is my question:

Why does the derivative s = f(x) = x^2 + 432/x determine the minimum amount of material needed to make the box..... what determines the maximum?

 

[JakeD]

Re: max and min problem

A rectangular box with a square base and no top is to have volume of 108 cubic inches. Find the dimensions for the box that require the least amount of material.

The function that is to be minimized is the surface area (s) while the volume (v) remains fixed at 108 cubic inches.

Letting x = length of the squre base and h = height of hte box you find that

V= x^2h = 108 cu in
h = 108/x^2

s= x^2 +4xh
s = f(x) = x^2 + 4x(108/x^2)
f(x) = x^2 + 432/x

with the domain of f(x) = (0, +infinity) because x represents a length.
f'(x) = 2x - 432/x^2
f'(x) = 0
2x- 432/x^2 = 0

2x^3 - 432 = 0
2x^3 = 432
x^3 = 216
x = 6

hence, a critical point occurs when x = 6. Using the 2nd derivative test:
f''(x) = 2 + 864/x^3
f''(6) = 6 > 0
and f has a local minimum at x = 6; hence, the dimensions of the box that require the least amount of material are a length and width of 6 inches and a height of 3 inches.

Okay, sorry for the long drawn out problem. But here is my question:

Why does the derivative s = f(x) = x^2 + 432/x determine the minimum amount of material needed to make the box..... what determines the maximum?

There is no maximum. The surface area f(x) = x^2 + 432/x is unbounded towards +infinity as the length and width x increase and the height h = 108/x^2 shrinks towards zero. It's hard for me to imagine, but apparently there is no limit to how much surface area can be wasted here enclosing a fixed volume.