Volume and max volume question?

So, the dimensions of the cardboard are determined by the expression:

2L + 2W = 13

W = 6.5 - L

Since we know that the creases will be evenly spaced, the length, x, of the distance between the creases will be:

x = L/4

L = 4x

We can put W in terms of x, also:

W = 6.5 - 4x

The box formed will have two sides of length x and one side of length W:

V = x²W = x²(6.5 - 4x)

V = 6.5x² - 4x³

Now, take the derivative and set it equal to zero to find the maximum:

V' = 13x - 12x² = 0

x(13 - 12x) = 0

x(x - 13/12) = 0

x = 0 or 13/12

Obviously, x cannot be 0. The maximum, then, occurs at:

x = 13/12 in.

The dimensions of the cardboard are:

L = 4(13/12) = 13/3 in.

W = 6.5 - L = 13/6 in.

P = 2L + 2W = 26/3 + 13/3 = 39/3 = 13 in.

The maximum volume is:

V = x²W

V = (13/12)²(13/6)

V = 2.54 in³.

This is impossible to answer with the information given as we have too many unknowns. We know the cardboard is rectangular, and that the perimeter is 13". So we have 4 sides that sum up to 13...

let a = the height and b = the length (such that our folds are parallel to side a)

we have 2a + 2b = 13, b = (13-2a)/2

Now we know that length b is to be divided into 4 sections by making 3 folds. Since the question asks for one value for x, we must assume that the folds are equally spaced (otherwise we would need a distance for each fold). However, we can assume that the maximum volume would be a cube with a square cross section anyway, so we need only 1 x, such that:

x + x + x + x = b, or b = 4x, x = b/4

so x = b/4 and b = (13-2a)/2 so:

x = (13-2a)/8

Our maximum volume would therefore be x*x*a or:

[a(13-2a)(13-2a)] / 64

Let

4x = width cardboard

y = length cardboard

P = perimeter cardboard

V = volume box

Given

P = 13

We have

P = 2*4x + 2y = 8x + 2y = 13

2y = 13 - 8x

y = 13/2 - 4x

V = x²y = x²(13/2 - 4x) = 13x²/2 - 4x³

Take the derivative and set equal to zero to find the critical values.

dV/dx = 13x - 12x² = 0

x(13 - 12x) = 0

x = 0, 13/12

Take the second derivative to find the nature of the critical values.

d²V/dx² = 13 - 24x

d²V/dx² = 13 - 0 = 13 > 0 for x = 0

This implies a relative minimum and is therefore rejected.

d²V/dx² = 13 - 24(13/2) = 13 - 156 = -143 < 0

This implies a relative maximum which is what we want.

x = 13/12

The maximum volume of the box is:

V = 13x²/2 - 4x³ = (13/2)(13/12)² - 4(13/12)³

V = 6(13/(2*6))(13/12)² - 4(13/12)³

V = 6(13/12)(13/12)² - 4(13/12)³ = 6(13/12)³ - 4(13/12)³

V = 2(13/12)³ ≈ 2.54 in³

Let the dimensions of the paper be x and (6.5 - x) inches.

This is to comply with the fact that the perimeter must be 13".

Now let the x dimension be the one which is divided by the three parallel lines. (You would get the same answer by using the other one but the working would be harder.)

Now it can be shown (but I will not bother as it is too easy) that the maximum area of the open end would be obtained by it being a square. therefore the lines divide the side equally into four, each part x/4 wide.

The volume produced is thus

V = (x/4)*(x/4)*(6.5 - x) = (6.5x^2 - x^3)/16.

Differentiation gives dV/dx = (13x - 3x^2)/16. Maximum volume will occur when dV/dx = 0 ---> 13x - 3x^2 = 0 ---> x = 0 or x = 13/3. Clearly x = 0 gives zero volume so we want 13/3.

x = 13/3 means length of box = 6.5 - 13/3 = 13/6

Therefore Vmax = (13/12)*(13/12)*(13/6) = 2197/864 = 2.54

Edit:

misread the number of creases

Let x = L/4

then 8x + 2W = 13

w = (13 - 8x)/2

and V = Wx^2 = (13x^2 - 8x^3)/2

dV/dx = 13x - 12x^2 = 0 for max volume

x(12x - 13) = 0

x = 0 is a trivial solution, so

x = 13/12 in

L = 13/3

W = 65/12

V ≈ 2.54 in^3

P=13=2*4x+y=8x+y

*note that x and y must each be greater than zero.

V=(x^2)*y

the bottom of the box is a square with side length x, and the box is y units tall.

y=13-8x

V=13x^2-8x^3..........Substitute and distrubute

V'=26x-24x^2...........Take the derivative

V'=0=2x(13-12x)......Set the derivative equal to zer and simplyfy

2x=0, x=0 13-12x=0, x=13/12

because x cannot be 0, x must equal 13/12

V=5.09

You can find the volume by multiplying the length, width, and height of your box. You'll get the maximum volume by making the box equally long and high.