Problem: An open box with dimensions 2 inches by 3 inches by 4 inches needs to be increases in size to hold five times as much material~?

The dimension are (2 in by 3 in by 4 in), so it gives us : 2 * 3 * 4 = 24 in³

Each dimension needs to be increases in size to hold five times as much material.

The new volume is: 5 * 24 = 120 in³



The new dimensions are:

The dimension 2 becomes: (2 + x)

The dimension 3 becomes: (3 + x)

The dimension 4 becomes: (4 + x)


…and the new volume is:

= (2 + x).(3 + x).(4 + x) → recall the new volume: 120

(2 + x).(3 + x).(4 + x) = 120

(6 + 2x + 3x + x²).(4 + x) = 120

(6 + 5x + x²).(4 + x) = 120

24 + 6x + 20x + 5x² + 4x² + x³ = 120

x³ + 9x² + 26x - 96 = 0 → it's necessary to eliminate the term at the power 2


x³ + 9x² + 26x - 96 = 0 → let: x = z - 3

(z - 3)³ + 9.(z - 3)² + 26.(z - 3) - 96 = 0

[(z - 3)².(z - 3)] + 9.(z² - 6z + 9) + 26z - 78 - 96 = 0

[(z² - 6z + 9).(z - 3)] + 9z² - 54z + 81 + 26z - 174 = 0

[z³ - 3z² - 6z² + 18z + 9z - 27] + 9z² - 28z - 93 = 0

[z³ - 9z² + 27z - 27] + 9z² - 28z - 93 = 0

z³ - 9z² + 27z - 27 + 9z² - 28z - 93 = 0

z³ - z - 120 = 0 ← you can see that there is no power 2

z³ - z - 120 = 0 → let: z = (u + v)

(u + v)³ - (u+ v) - 120 = 0

[(u + v)².(u + v)] - (u+ v) - 120 = 0

[(u² + 2uv + v²).(u + v)] - (u+ v) - 120 = 0

[u³ + u²v + 2u²v + 2uv² + uv² + v³] - (u+ v) - 120 = 0

[u³ + v³ + 3u²v + 3uv²] - (u+ v) - 120 = 0

[(u³ + v³) + (3u²v + 3uv²)] - (u+ v) - 120 = 0

[(u³ + v³) + 3uv.(u + v)] - (u+ v) - 120 = 0

(u³ + v³) + 3uv.(u + v) - (u+ v) - 120 = 0 → you can factorize (u + v)

(u³ + v³) + (u + v).(3uv - 1) - 120 = 0 → suppose that: (3uv - 1) = 0 ← equation (1)

(u³ + v³) + (u + v).(0) - 120 = 0

(u³ + v³) - 120 = 0 ← equation (2)


You can get a system with 2 equations:

(1) : (3uv - 1) = 0

(1) : 3uv = 1

(1) : uv = 1/3

(1) : u³v³ = 1/27


(2) : (u³ + v³) - 120 = 0

(2) : u³ + v³ = 120


Let: U = u³

Let: V = v³


You get a new system:

(1) : UV = 1/27 ← this is the product P

(2) : U + V = 120 ← this is the sum S


…and you know that U et V are the solutions of the following equation: x² - Sx + P = 0

x² - 120x + (1/27) = 0


Δ = 120² - [4 * (1/27)]

Δ = 14400 - (4/27)

Δ = (388800/27) - (4/27)

Δ = 388896/27

Δ = (74² * 71) / (3² * 3)

Δ = (74/3)² * (71/3)


x1 = [120 - (74/3).√(71/3)] / 2 = 60 - (37/3).√(71/3) ← this is U

x2 = [120 + (74/3).√(71/3)] / 2 = 60 + (37/3).√(71/3) ← this is V


Recall: u³ = U

u³ = 60 - (37/3).√(71/3)

u = [60 - (37/3).√(71/3)]^(1/3)


Recall: v³ = V

v³ = 60 + (37/3).√(71/3)

v = [60 + (37/3).√(71/3)]^(1/3)


Recall: z = (u + v)

Recall: x = z - 3


x = (u + v) - 3

x = [60 - (37/3).√(71/3)]^(1/3 + [60 + (37/3).√(71/3)]^(1/3) - 3

x = 2 ← this is the increase

So the old box has a volume = 2 * 3 * 4 = 24 cubic inches.
You want the new box to have a volume of 24 * 5 = 120 cubic inches.

You are increasing each dimension by the same amount, let's call the increase (x) inches to get the new volume (V). {Do you mean the same amount, or the same fraction? I'm assuming the same amount!}

So:

(2 + x) * (3 + x) * (4 + x) = V
Multiply out the brackets. Let's do it one step at once.
(6 + 2x + 3x + x²)(4 + x) = V
(6 + 5x + x²)(4 + x) = V
24 + 6x + 20x + 5x² + 4x² + x³ = V

V = x³ + 9x² + 26x + 24

In this case V = 120
120 = x³ + 9x² + 26x + 24
x³ + 9x² + 26x - 96 = 0

Solving cubic equations by hand is a grisly business, so I'm off to an on-line solver:
http://www.wolframalpha.com/widgets/view.jsp?id=3f4366aeb9c157cf9a30c90693eafc55

Which gives us one real root at (x = 2)
Giving the new box the dimensions:
(2 + 2)(3 + 2)(4 + 2)

4 * 5 * 6 = 120 cubic inches.

The current volume is (2)(3)(4) = 24
New volume = (24)(5) = 120
Let x = amount added to each side
New Dimensions are (x + 2)(x + 3)(x + 4) = 120
(x + 2)(x^2 + 7x + 12) - 20 = 0

This gets you set up and going. YOU need to finish the multiplication, that will give you "the function that represents the volume of the new box"

Next, solve for x.

The other answers are correct, they just didn't show you how.

1.
V : volume of the box in cubic inches.

It is assumed each dimension is increased by the same amount.
Let the new dimensions be 2y inches × 3y inches × 4y inches

Old volume = 2 × 3 × 4 cubic inches
New volume = 2y × 3y × 4y cubic inches = 24y³ cubic inches

The function : V = 24y³


2.
Volume is increased 5 times :
2y × 3y × 4y = (2 × 3 × 4) × 5
y³ = 5
y = ³√5

New dimensions : 2(³√5) inches × 3(³√5) inches × 4(³√5) inches

All dimensions are to be increased by factor 5^(1/3) ~ 1.71

Volume is proportional to length^3

So length is proportional to Volume^(1/3)