Find dimensions- open top box, square base, must hold 32 cubic inches, built w/minimum materials?

Question

Basically, an open-topped box with a square base. The box must hold 32 cubic inches. Find the dimensions of the box that can be built with minimum amount of materials.

Answer 1

Let
a = length of side on base of box
h = height box
V = volume box
S = surface area box

Given
V = 32

Find a and h to minimize S.

V = a²h
h = V/a²

S = a² + 4ah = a² + 4a(V/a²) = a² + 4V/a

To find the critical points take the derivative of S and set it equal to zero.

dS/da = 2a - 4V/a² = 0
2a = 4V/a²
a = 2V/a²
a³ = 2V
a = (2V)^(1/3)

Take the second derivative to determine the nature of the critical point.

d²S/da² = 2 + 8V/a³ > 0 for all a
So it is a relative minimum which is what we want.

Plug into h.

h = V/a² = (2V/a²)/2 = a/2

Now plug into S.

S = a² + 4ah = a² + 4a(a/2) = a² + 2a² = 3a²
S = 3[(2V)^(1/3)]² = 3(2V)^(2/3)
S = 3(2*32)^(2/3) = 3*64^(2/3) = 3*16 = 48 sq in

Answer 2

To solve this min/max problem, we first require the formula for the volume of this box. Normally, the volume of a box is equal to length times width times height, but in this case, length = width. That makes the formula

V = (x^2)h

But V = 32, so

32 = (x^2)h

We also require the surface area of this box. Since there is no top, the surface area should be

A = (area of the bottom) + (area of the 4 walls)
A = x^2 + 4xh

But, from the first equation, we can solve for h.
32 = (x^2)h implies that h = 32/(x^2). Plug this value of h into our area equation, and we get

A = x^2 + 4x[32/(x^2)]

Simplifying, we get

A = x^2 + 4(32/x)
A = x^2 + 128/x

Making into a single fraction,

A = (x^3 + 128)/x

Now that we have one variable, we will now declare this as our area function, A(x).

A(x) = (x^3 + 128)/x

To minimize the area, take the derivative A'(x) and then make it 0. Using the quotient rule,

A'(x) = [ (3x^2)(x) - (x^3 + 128)(1) ] / (x^2)
A'(x) = [ 3x^3 - x^3 - 128 ] / (x^2)
A'(x) = [ 2x^3 - 128 ] / (x^2)
A'(x) = 2(x^3 - 64) / (x^2)

Make A'(x) = 0,

0 = 2(x^3 - 64) / (x^2)

We can neglect the bottom, because x^2 is not equal to 0 (x must represent a dimension).

0 = 2(x^3 - 64)
0 = x^3 - 64
64 = x^3

Therefore, x = 4

The minimum area occurs at x = 4.
We can derive h, since
h = 32/(x^2)
h = 32/(4^2)
h = 32/16
h = 2

The dimensions of the box that can be bult with minimum amount of materials would be

4" x 4" x 2"

As a side note, the minimum surface area would be A(4), or
A(4) = (4^3 + 128)/4
A(4) = (64 + 128)/4
A(4) = 192/4
A(4) = 48 square inches

Answer 3

Here is a solution using an Excel Spreadsheet.
L W H Lateral Surface Area
32/(L)(W) [(L)(W) + (4)(L)(H)]
1 1 32.00 129.00
2 2 8.00 68.00
3 3 3.56 51.67
4 4 2.00 48.00
5 5 1.28 50.60
6 6 0.89 57.33
7 7 0.65 67.29
8 8 0.50 80.00
9 9 0.40 95.22
10 10 0.32 112.80
11 11 0.26 132.64
12 12 0.22 154.67
13 13 0.19 178.85
14 14 0.16 205.14
15 15 0.14 233.53
ANSWER:
Requirement:
Minimum Lateral Surface Area = 48
Dimension:
Length = 4 inches
Width = 4 inches
Height = 2 inches
Remark:
No Calculus or Algebraic equations were used.
Only Mathematical formulas were used in the cells.

Answer 4

Materials required will follow this formula (L = length, W = width, H = height)
L = W
Material = L*L + 4 * L * H
M = L^2 + 4LH

V = L*L*H = 32
H = 32/L^2

M = L^2 + 4L(32/L^2)

Differentiate and find the zero of the equation

L = W = 4
H = 2

Volume = 4 * 4 * 2 = 32
M = 4 * 4 + 4 * 4 * 2 = 16 + 32 = 48

Answer 5

500 Cubic Inches

Answer 6

base of side x and height h

volume,
x^2 * h = 32

h = 32 /( x^2 )------(1)

suface area S = x^2 + 4xh
using (1)

S = x^2 + 128 / x

dS/dx = 2x -128 / (x^2)

S'' = 2 +256 / x^3

for stationary values dS/dx = 0
2x -128 / (x^2) = 0 implies x = 4
S"(4) >0

S is min. for x = 4
and h = 32 /( x^2 ) = 2

hence 4 X 4 X 2

Answer 7

To minimize the materials the box should be in the shape of a cube. Each side should be the cube root of 32,000 cm3 or approx 31.75 cm. area of materials is 31.75^2*5 sides = approx 5040.3 cm^2 try any other dimensions that come to 32000cm^3 and the area will be larger.