# Use a standard A4 paper to form a largest volume block

• May 14th 2006, 06:28 AM
abcting223hk
Use a standard A4 paper to form a largest volume block
how to Use a standard A4 paper to form a largest volume block ?
• May 14th 2006, 10:55 AM
earboth
Quote:

Originally Posted by abcting223hk
how to Use a standard A4 paper to form a largest volume block ?

Hello,

you have to cut sqares from the corners of the sheet of paper. I've attached a diagram to demonstrate what I've calculated.

Let be l = length of sheet, w = width of sheet. Then you get:

$V=(l-2x)(w-2x) \cdot x=4x^3-2x^2(l+w)+x(l+w)$

x must be lower then .5*w.

You'll get the maximum of V if the 1rst dervative of V with respect to x equals zero:

${dV \over dx}=12x^2-4x(l+w)+(l+w)$

$0=12x^2-4x(l+w)+(l+w)$. Solve for x. You'll get 2 solutions, but the greater solution isn't very realistic.

$x={1\over 6} \cdot \left(w+l-\sqrt{w^2-w \cdot l+l^2} \right)$. That will give approximately x = 4.04 cm.

Plug in w = 21 cm and l = 29,7 cm.

Greetings

EB
• May 14th 2006, 12:02 PM
CaptainBlack
Quote:

Originally Posted by earboth
Hello,

you have to cut sqares from the corners of the sheet of paper. I've attached a diagram to demonstrate what I've calculated.

Let be l = length of sheet, w = width of sheet. Then you get:

$V=(l-2x)(w-2x) \cdot x=4x^3-2x^2(l+w)+x(l+w)$

x must be lower then .5*w.

You'll get the maximum of V if the 1rst dervative of V with respect to x equals zero:

${dV \over dx}=12x^2-4x(l+w)+(l+w)$

$0=12x^2-4x(l+w)+(l+w)$. Solve for x. You'll get 2 solutions, but the greater solution isn't very realistic.

$x={1\over 6} \cdot \left(w+l-\sqrt{w^2-w \cdot l+l^2} \right)$. That will give approximately x = 4.04 cm.

Plug in w = 21 cm and l = 29,7 cm.

Greetings

EB

I took this to mean: what is the block with largest volume that can be
wrapped with a sheet of A4?

RonL
• May 15th 2006, 02:19 AM
earboth
Quote:

Originally Posted by abcting223hk
how to Use a standard A4 paper to form a largest volume block ?

Hello,

according to CaptainBlack's hint (thanks a lot, I wouldn't have seen it this way) you have to cut three couples of sqares from the sheet of paper.

Let be l = length of sheet, w = width of sheet and x = half the height of the block. Then you get:

$V={(l-3x)(w-2x) \over 2}\cdot 2x=6x^3-x^2(2l+3w)+x(l*w)$

x must be lower then .5*w.

You'll get the maximum of V if the 1rst dervative of V with respect to x equals zero:

${dV \over dx}=18x^2-2x(2l+3w)+(l*w)$

$0=18x^2-2x(2l+3w)+(l*w)$. Solve for x. You'll get 2 solutions, but the greater solution isn't very realistic.

That will give approximately x = 3.3956 cm.

Greetings

EB
.
• May 15th 2006, 06:51 AM
abcting223hk
I am a form 3 student from HK.,it is really difficult for me to understander the answer.Could you help me?
• May 15th 2006, 09:01 PM
earboth
Quote:

Originally Posted by abcting223hk
I am a form 3 student from HK.,it is really difficult for me to understander the answer.Could you help me?

Hello,

maybe it's easier for you if I use numbers instead of variables.
The A4-format of a sheet of paper means that this sheet is 29,7 cm long and 21,0 cm wide.

If you wrap a block into a sheet of paper then you need a margin strip of paper which has a width of half of the block's height to cover the block from above and from the bottom. These strips are made by cutting off little squares which sides are as long as the half of the block's height. Please look at the attached diagram.

The volume of a block is calculated by
$V=\mbox{length of block} \cdot\mbox{width of block} \cdot\mbox{height of block}$

That means you have to compute the length of the block = 21 cm - 2x,
the width of the block = (29.7 cm - 3x) / 2 because you have to calculate the bottom and the cover of the block simultanously(?),
the height of the block = x.

You'll get a function in x:

$
V(x)={(29.7-3x)(21-2x) \over 2}\cdot 2x$
= $6x^3-x^2(2 \cdot 29.7+3 \cdot 21)+x(29.7*21)$

$V(x)=6x^3-122.4 \cdot x^2+623.7 \cdot x$

x ranges from 0(zero) to 10.5 cm (that is half of the paper's width)

You'll get the maximum of V if the 1rst dervative of V with respect to x equals zero:

${dV \over dx}=18x^2-244.8x+623.7$

So you have to solve the equation:

$0=18x^2-244.8x+623.7$

This is a quadratic equation which has the solutions:

$x_1 \approx 10.2044\ \ \vee \ \ x_1 \approx 3.3956$

Remember please: x is the half of the block's height.

The first solution isn't very realistic because 3x is greater than the length of the paper.

Greetings

EB
• May 16th 2006, 04:36 AM
abcting223hk
why dV/dx=18x^2-244.8x+623.7?
why x1 =approx 10.2044 V x1 =approx 3.3956?
• May 16th 2006, 06:54 AM
earboth
Quote:

Originally Posted by abcting223hk
why dV/dx=18x^2-244.8x+623.7?
why x1 =approx 10.2044 V x1 =approx 3.3956?

Hello,

1. What is 3x?
Please look at the diagram which was attached the previous reply. You can see, that I took off 3 times the side of the little square. The lengt of the side of one square is x, because I don't know the actual value. And now you've got left (29.7-3x) to wrap a block in.

2. why dV/dx=18x^2-244.8x+623.7?
Do you know to calculate the maximum or the minimum of a function? You have to calculate the first derivative (which will give you the value of the gradient of a function) and set this equal zero, because at a maximum or at a minimum the gradient has the value zero.
If you are not familiar with derivatives of function, I'm not quite sure that I could help you.

3. why x1 =approx 10.2044 V x1 =approx 3.3956?
As you've seen I solved a quadratic equation which may have none, exactly one or two solutions. With this equation there are two solutions which I named x1 and x2. So the 1 and the 2 are indices of x.

Greetings

EB

(By the way: What does HK mean?)
• May 16th 2006, 02:42 PM
abcting223hk
HK=hong kong
• May 17th 2006, 02:20 AM
abcting223hk
Why the height is 2X?
• May 17th 2006, 08:40 PM
earboth
Quote:

Originally Posted by abcting223hk
Why the height is 2X?

Hello,