one more word problem (calculus) can someone please help?

Question

can anyone help me with this please? I've been able to get them all except for this word problem and the one other which I posted :[

A television manufacturing firm needs to design an open-topped box with a square base. The box must hold 32 in^3 Find the dimensions of the box that can be built with the minimum number of materials. (then there's a diagram of what the box looks like. It is a cube with 4 sides and a bottom. There is no top.)

Answer 1

The base is square, so let the base side length be x in. Then the base area is x^2 and the volume is 32 in^3, so the height must be h = 32/x^2 in.
The amount of materials required corresponds to the surface area, so we get
A = x^2 + 4xh
= x^2 + 4(32/x^2)
= x^2 + 128/x
dA/dx = 2x - 128/x^2
= 2(x^3-64) / x^2
= 0 <=> x^3 = 64 <=> x = 4 in.
d2A/dx2 = 2 + 256/x^3 > 0 for x > 0, so this is indeed a minimum.
So the base must be 4 in on a side, and the height is 2 in.

Answer 2

Let's call the dimensions of the box x by x by h. Therefore, its volume is x^2h which we are told equals 32.

x^2h = 32
h = 32/x^2

The surface area of a box with no top is:
A = x^2 + 4xh
A = x^2 + 4x(32/x^2)
A = x^2 + 128x^-1

A' = 2x - 128x^-2
0 = 2x^-2(x^3 - 64)
critical values are 0 and 4, but 0 is not a valid value for x since it represents a length of a side of the box. But, you should still test that 4 is actually where the minimum occurs.

0 - - - - - - - 4 + + + + + +

Yes, 4 is a minimum of the area function, which would minimize the amount of material. So the dimensions are 4 x 4 x 32/4^2 = 4 x 4 x 2

Answer 3

volume = a^2 b = 32

total materials = area of bottom + area of the sides
total materials = m = a^2 + 4 a b

Express in terms of a
m = a^2 + 4 a (32/a^2)
m = a^2 + 128/a
m = a^2 + 128a^(-1)
Find the derivative dm/da
dm/da = 2a - 128a^(-2)

When m is minimized, dm/da = 0

2a = 128 a^(-2)

2a^3 = 128

a^3 = 64

a = 4

So, for the minimum materials for box of volume 32,
the base is 4 x 4 inches
the height is 2 inches