Open box Problem.

6X=L

For example, when the length is 24cm,L=24cm, we get 6X=24cm

6X=24

X=24/6

X= 4

Then max. value of X will equal to 4cm when the length is 24cm.

        When the length is 12cm,L=12cm, we get 6X=12cm

        6X=12

X=12/6

X= 2

We have found that these results are correct.

And we can see that they have the same proportion, so that I am sure that the formula will work on any size.

 

Then, we get a formula to calculate the max. value of X for any size of a square, because the value of X is proportional to the value of  L, as they are similar shapes. As the value of X increase, L will also increase in proportional, so that you can use this formula to calculate the max. value of X.

X = L/6

From this formula, we can find out the formula to calculate the max. volume for any size of square.

We put X= L/6 into this formula X(L-2X), then we will get (L-), after we simplifying this, we get this formula:                  to calculate the maximum volume of a square in any size!!

Now I will test the formula

Using the size 12cm to the formula         

= 128

At last we can see that the formula is correct.

Open box problem 2

        Now I am going to find out the size of an open box with 2 different of lengths. We know that there are too many combinations of lengths and widths, Now I am going to try a card with the size of Wx2W, then Wx3W, and Wx10W and see if I can find a connection with the results. And all the calculations were calculated by using spreadsheets from Microsoft Excel.

We know that the height is still be X.

Then the equation for one of the side is         W – X – X.

And the other one is         2W – X – X.

The formula to calculate the volume of the open box is (2W – X –X )x(W – X - X)xX.

First I am going to set the value of W be 12 cm, so that 2W will be 24cm, this is a ratio of 1:2, the length being twice as long as the width. Here are the results:

I found out that the maximum value of X to get the maximum volume is 3. In order to further confirm the finding,  I decided to try value of X between 2.1cm and 3cm.

Here are the results:

 

We can see that 2.5cm is the maximum value of X. To get further confirmation of the result, I will try X between 2.41cm and 2.6cm.

We can see that the maximum value of X is 2.54cm

Now I am going to try between 2.535 – 2.545

We can see that the max. value of X is 2.536cm, that means 2.54cm correct to 2 decimal places.

Then we divide the width by X, 12/2.54, we get      4.724

I expect this will be the result for all rectangles in the same ratio.

Now I am going to try W=6cm and 2W=12cm, with also the same ratio 1:2.

With a cut out of 1cm,X=1cm, 2cm is the maximum for there to be a box left.

We can see that the maximum value of X is 1cm

I will try the value X  between 1 and 2.

1.3cm  is the maximum value of X. To further confirm the result, I will try X  between 1.21cm and 1.3cm.

From the results we can see, 1.27cm is the maximum value of X.

Now I am going to try between 1.265 and 1.275.

We can see that the max. value of X is 1.268cm, that means the value is 1.27cm correct to 2 decimal places.

Then we again divide the width by X, 6/1.27, we get                 4.7244

So we get a formula to calculate the max. value of X for a 1:2 rectangle, because the value of X is proportional to the value of  W, as they are similar shapes. As the value of X increase, W will also increase in proportional, so that you can use this formula to calculate the max. value of X in a 1:2 rectangle.

X=W/4.7244

Now I am going to set the length at a ratio of 1:3, the width will be 12cm and the length will be 36cm.

With a cut out of 1cm,X=1cm, 5cm is the maximum for there to be a box left.

Here are the results:

We can see that 3cm is the biggest value for the volume, In order to further confirm the result, I will try between 2.1 and 3.

We can see that 2.7cm is the biggest value, now I will try between 2.65 and 2.75

We find that 2.71cm is the biggest value for the maximum volume of the box.

Now I will try between 2.705 and 2.715

We can see that 2.709cm is the maximum value of X, the value of X will be 2.71cm correct to 2 decimal places.

Then we divide the width by the value of X, 12/2.71, then we get 4.428…..

As I say in 1:2 rectangle, the value of X is proportional to the value of  W, as they are similar shapes. As the value of X increase, W will also increase in proportional, so that I expect the ratio 1:3 will also has the same, so get this formula to calculate the max. value of X in a 1:3 rectangle.

X=W/4.428

Now I will set the width of 12cm and the length 120cm, at the ratio of 1:10

With a cut out of 1cm,X=1cm, 5cm is the maximum for there to be a box left.

We can see that 3cm is the biggest value. To further confirm the result, I will try between 2.5cm and 3.4cm.

From the result we can see 2.9cm is the biggest value, now I will try between 2.85cm and 2.94cm.

We can see 2.92cm is the biggest value for the maximum volume of the box.

Now I am going to try between 2.915 and 2.925

We can see 2.92cm is the biggest value for the maximum volume of the box.

We then divide the width by X, 12cm/2.92cm=        4.10958

As I say above, for this ratio 1:10 of a rectangle, I will expect the formula will be

X=W/4.10958

As the ratio of sides increase, the value of the width/max. value will decrease, as the ratio is bigger, the value will get closer to 4, as I predicted on the graph, so we assume that the value is 4, then we get this formula:

X=W/4

This is the formula to calculate the approximate maximum value of X for a rectangle. So if the ratio get bigger, you can get a approximate value of X

The formula for calculating the box is:

V=X(W-X-X)(L-X-X)

We can simplify this formula like that:

V=X(W-2X)(L-2X)

V=(XW-2X)(L-2X)

V=XWL- 2XL- 2XW +4X

We put  X= into this formula, we will get:

V=()WL- 2()L- 2()W +4()

After simplifying it, and then we will get the final formula:

Here is the formula to calculate the maximum volume of a rectangle box. But this formula can only find out the approximate volume of the box. As the ratio get bigger, you can get a more accurate result. Because the value of X is not the same on different ratio of rectangle and the value is getting nearer to 4 when at a bigger ratio. By this formula, you can very easily find out the approximate maximum volume of a rectangle box.