Math Question What are the Dimensions for a box that require...?

Question

A Box with an open top and a square base is to be constructed to contain 4000 cubic inches. Find the dimensions for the box that will require the minimum amount of material to construct the box. What are the dimensions? (please show me how to do this thanks)

Answer 1

Let x be the length of one side of the square base.
Let y be the height of the box.

We know x*x*y = 4000
From this, we get y = 4000 / (x^2)

We want to minimize (x*x) + 4*(x*y)
Plugging in y, we get:
x^2 + 4*x*(4000/x^2)
That's x^2 + 16000/x

To find where this equation is at its minimum, take the derivative and set it to zero.

2x - 16000/(x^2) = 0
Multiplying through by x^2:
2x^3 - 16000 = 0
2x^3 = 16000
x^3 = 8000
x = 20

Plugging this into the equation for y:
y = 4000/400 = 10

The dimensions of the box are 20*20*10 inches.

The preceding answer is wrong, "Dave" accidentally doubled 16000 to 32000 for no reason.

Answer 2

The key here is that you want a box that has minimum materials. Since the side opposite the base (the top) is missing, and is the same dimensions as the base; you want the top to be as large as possible.

So make the base as large as possible.

Make the base 4000 inches square, and the sides 1 inch high.

Answer 3

Assuming metal is very thin

base is square let the base be l
let height be h
area = l^2 + 4hl
and volume = (l^2)h = V
=> h = V/l^2
A = l^2 + 4hl = l^2 + 4V/l
For min A , A' = 0
=> 2l - 4V/l^2 = 0
=> l^3 = 2V
l = CubeRoot(2V) = CubeRoot(8000) = 20 inch

Answer l = 20 inch , h = 10 inch

Answer 4

a finished sum of 60" must be 20+20+20. this might provide you a quantity of 20 x 20 x 20 = 8000 cu in. If 10+30+20 = 60" this supplies a quantity of 10 x 30 x 20 = 6000 cu in. etc. think of the cube might provide you greater quantity.

Answer 5

Volume = d^2 x h = 4000
Area = d^2 + 4dh
To minimize area, take derivative with respect to d and set to zero.
2d+4h = 0
therefore d=-2h
Substitute into volume equation:
(-2h)^2 x h = 4000
4h^2 x h = 4000
h^3 = 1000
h = 10
and substituting again
d^2 x 10 = 4000
d^2 = 400
d = 20
20" x 20" x 10"

Answer 6

l*w*h = 4000

l = w { because it has a square base }

w²h = 4000
h = 4000/w²

materials = w² + 4wh = w² + 4*w*4000/w² = w² + 16000/w

Differentiate this equation, and set it equal to zero to find the minimum:

dw = 2w - 16000/w² = 0

2w = 16000/w²

2w³ = 32000

w³ = 16000

w = ³√(16000) = 25.198420997897463295344212145565

h = 4000/w² = 6.2996052494743658238360530363911