Open box Problem.

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Open box Problem

Introduction

        Now I am going to investigate the size of the cut out square which makes an open box of the largest volume. Let the length of the square be 24cm, the card is shown as below:

                                                           

Method

Now, I am going to set the length of the square to be 24cm and the other one will be 12cm. Then I will find out the relation between the two experiment and find out a formula to calculate the max. volume for any size of square. Then I will explain why the formula works for all squares.

I have to find out the value of X to give the maximum of volume.

First, I have to find out the formula to calculate the volume of the open box. Base on the formula, the volume= height x length x width.

Let the height be X.

Then the length is 24cm – X- X .

Therefore we get  X(24cm-2X)

With a cut out of 1cm,X=1cm, 11cm is the maximum for there to be a box left.

Here are the results:

We found out that when X=4, we can get the maximum volume in integer.

Now I am going to find out the volume when X is between 3.5 and 4.5.

Here are the results:

We found out that 4cm is still the biggest value whereas we can get the biggest volume.

Now I am going to set the length to be 12.

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

Then we get the following results:

We found that 2 is the biggest value whereas we can get the biggest volume in integer.

To get a more accurate result, we have to try 1 decimal place. Now I am going to set the value of X between 1.5 and 2.5.

Here are the results:

We found out that 2cm is still the biggest value.

Conclusion

From the above 2 results we find out that when the length is 24cm, the value of X for the max. volume is 4cm. When the length is 12cm, then the value of X for the max. volume is 2cm. And there’s a relation between these 2 results, we find out that the length of the card is equal to 6 times of the max. value of X.

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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 ...

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