Example:
x = 2 20X20 Square
v = x(l-2x)(w-2x)
v = 2(20-4)(20-4)
v = 2*16*16
v = 512cm³
Now I will do a trail run with a 20X20 square. The results will be given on the table below:
After the trail run, I found out that length 3 gives the highest volume, and the volume started to get lower when the small square’s length gets higher after 3, therefore my prediction is wrong. According to the graph, I saw that there should be higher volumes in between 3 and 4, so now I use ratio to find the formula. The only ratio that would fit is 1:6, since 1/6 of 20 is 3.3. I would have to prove that this formula/ratio should work on every sized square box.
Now I will try a 30X30 square box:
In the 30X30 square box, 5 give the highest volume. This shows that the ratio of 1:6 (5:30) also work on a 30X30 square box. Although it worked on both 20X20 squares and 30X30 squares, it might still be wrong. So therefore I will try a 40X40 box to make sure.
A 40X40 square table:
In a 40X40 square, the largest volume shown in the table is 7. By using the ratio 1:6, I get 6.7 as the result. Now I will do a graph in order to prove that the largest volume should lie on 6.7.
According to my graph, the point between 6 and 7 went higher than 7. This is an evidence for the ratio of 1:6 being true.
To prove it even further, I will use the rule of differentiation to prove this formula. Differentiation is used since it shows the rate of change of a curve, which in this case would be the volume curve: x (l-2x)(w-2x)
Multiplied out the volume formula is 4x³-2x²w-2x²l+xlw
n-1
To calculate x, newton’s law of differentiation dv/dx (d=change in) = nx is used. I will first have to differentiate the formula of 4x³-2x²w-2x²l+xlw. This formula would then become:
12x²-4xw-4xl+lw
Now I will use the 20X20 square to work out the formula, first by substituting the formula.
12x²-4x20-4x20+20*20
=12x²-80x-80x+400
=12x²-160x+400
I then equal the formula to 0. I do this since the gradient on the maximum rate of change is 0. My aim is to find the maximum volume. To solve x, I have to use the quadratic formula, since this quadratic equation cannot be factorised.
a =12 b =-160 c=400
x = -b +- √b²-4ac
2a
x = 160+- √160²-4*12*400
2*12
x = 160+- √25600-19200
24
x = 160+- √6400
24
x = 160+- 80
24
x = 160+ 80 x = 160-80
- 24
x = 240 x = 80
- 24
x = 10 x = 3.3
There are two answers given because not only the gradient of the maximum rate of change is 0, but also the minimum rate of change.
Out of these answers, 10 would not be the right answer since for a 20X20 square, if you choose 10 to be the small square’s length, you would not get a box, so the volume is 0, the minimum rate. (10+10=20) So 10 is the minimum, then 3.3 should be the maximum. As I had experimented, for a 20X20 square the largest volume lies on 3.3, so this proved that the ratio 1:6 works.
Now I would move to find the size of the squares cut for a rectangle to get the largest volume possible. For rectangles, I am going to decide the dimension of it, since the width and length are now different. I’m going to use the ratio 1:2 to work with. (For example: 10X20,15X30)
The Rectangle Open Box:
To find the volume of a rectangle box, I realize that I can use the same volume formula used for the square:
Volume = x (l-2x)(w-2x)
Example:
x = 2 l = 20 w =10
v = 2 (20-4)(10-4)
v = 2*16*6
v = 192cm³
Now I will do a trail run with a 20X10 rectangle. The results will be given on the table below:
On length 5 the volume drops to 0. This happens because the width is only 10 cm long. If you cut it more than 5cm, there would not be a side to fold and to form a rectangle box. For a 20x10 rectangle, shown in the table the largest volume lies on length 2. I didn’t find anything here yet, so I’m trying a 15X30 rectangle, which is still in ratio of 1:2.
Here is the table for a 15X30 rectangle:
For the 15X30 rectangle, the largest volume lies on length 3. I still don’t really see a pattern or ratio yet, so I’m going to do a 20X40 rectangle.
The table for 20X40 rectangle:
By reading the results, I see a pattern for the rectangles.
10X20: 2
15X30: 3
20X40: 4
I predict that for a 25X50 rectangle the largest volume will lay on length 5, because of the pattern I see now. So I’m going to try a 25X50 rectangle.
Table of 25X50 rectangle:
