15cm (15cm – 2x)
x cm
(24cm -2x)
Question 2a) For a rectangle with a length of 24cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):
= Length * Width * Height
= (24 – 2x) * (15 – 2x) * x
= 360x – 78x2 + 4x3
Essentially differentiating is done by taking an equation:
xn and applying the principle nx(n-1)
Therefore if: y = 360x – 78x2 + 4x3 =>dy= 360 – 156x + 12x2
dx
resulting in a quadratic equation, which can be solved by:
x = -b + or - √b2 – 4ac hence x = - (-156) + or - √ (156)2 – 4 (360)(12)
2a 2(12)
x = 156 + or - √ 24336 - 17280
24
x = 10cm and 3cm
Clearly the cutout size cannot be 10cm since the rectangular box is only 15cm wide. The only other possible cutout dimension would be 3cm.
To make sure my calculations are correct, I have drawn up a spreadsheet below with a graph showing me how increasing the cutout size affects the volume. As you can see from the table and the graph on the next page, the optimum cutout size for a rectangular box with a length of 24cm and a width of 15cm is 3cm.
In this spreadsheet, a number of negative numbers occur, however, these are not valid because it is impossible to obtain a negative value of length or width for any shape so I have decided to make this spreadsheet more sophisticated by incorporating an IF formula which will tell me when one of the dimensions of the rectangular box is negative.
I can generate an IF formula by using a relational operator with the value 0 as a way of representing negative numbers, for example, using the less than sign (<), I can make a statement saying that any number < 0 is negative, which is true. Either a message or a number can also be attached on to this in order to see whether the contents of a certain cell will have a certain word in them. However, if the values are messages, then they need to be enclosed inside double quote marks so that they separate the different parts within the brackets.
Using this information, I predict that if I key in the formula
=IF(B1-2*A5<0,"Error",B1-2*A5) into cell C5 nothing will happen because 15 – ( 2 x 0.0 ) = 15. The outcome of this calculation is greater than 0 so in theory no error sign should come up, instead the answer to the same calculation 15 – ( 2 x 0.0 ) should appear in cell C5. However, if I keyed in the formula =IF(B1-2*A21<0,"Error",B1-2*A21) into cell C21 an error sign should occur because 15 – ( 2 x 8 ) = -1. The outcome of this calculation is less than 0 so in theory an error sign should come up.
As you can see from the spreadsheet shown above, my prediction was right, however, in cells E21–E28 a ‘#VALUE!’ sign has appeared because the various volumes cannot be calculated as there are no values for the width of the box in cells C21–C28.
I will continue my investigation by examining other sizes of rectangles.
23 cm
x cm
15 cm (15 cm – 2x)
x cm
(23 cm –2x)
Question 2b) Similarly for a rectangle with a length of 23cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):
= Length * Width * Height
= (23 – 2x) * (15 – 2x) * x
= 345x – 76x2 + 4x3
Essentially differentiating is done by taking an equation:
xn and applying the principle nx(n-1)
Therefore if: y = 345x – 76x2 + 4x3 =>dy= 345 – 152x + 12x2
dx
resulting in a quadratic equation, which can be solved by:
x = -b + or - √b2 – 4ac hence x = - (-152) + or - √ (152)2 – 4 (345)(12)
2a 2(12)
x = 152 + or - √ 23104 - 16560
24
x = 9.704cm (4s.f) and 2.963cm (4s.f)
Clearly the cutout size cannot be 9.704cm (4s.f) since the rectangular box is only 15cm wide. The only other possible cutout dimension would be 2.963cm (4s.f).
Instead of creating a new spreadsheet, I can change values within an old spreadsheet to suit the new condition I want.
For example in the above spreadsheet, I predict that if I change cell B2 to 23 cell B5 and the rest of the column going downwards will change subtracting 1 each time from the value because of the formula =B2-2*A5. This formula adds the contents of cell B2 into cell B5 but it also subtracts 2 multiplied by the cutout size on the same row, so in this case the cutout size is 0.0 and for this reason the whole column changes because all the cells in the column B use a similar formula except for the final value which changes according to the row you are on. Naturally the volume figures will all change because you have changed the width values which affect the volume. In this case I predict that all the volume numbers will decrease when you change cell B2 to 23 except for the figures in cells E5 and E20 where both should remain at 0 because you have multiplied by zero to get the volume size. It should be noted that it is only valid to change the value in cell B2 rather than B1 because I am investigating the change in length of the box each time and not the width.
As you can see, my prediction regarding column B was correct, however, my prediction stating that all the volumes should decrease in size when you change cell B2 to 23 except for cell E5 and E20 was wrong because all the negative volumes increased in size.
Using the spreadsheet and the graph above, you can see that the optimum cutout size for a rectangular box with a length of 23cm and a width of 15cm is roughly 3cm.
My next prediction is that as the value of the length gets closer to the constant value of the width (15cm), the optimum cutout size decreases in size. In order to assess this prediction, more results need to be obtained from different lengths of rectangles where the width is always kept at 15cm. I have decided to keep one of the values constant whilst investigating because it is hard to draw any suitable conclusion from my evidence when both factors are constantly changing as the proof becomes almost void:
22cm
x cm
15cm (15cm – 2x)
x cm
(22cm – 2x)
Question 2c) Similarly for a rectangle with a length of 22cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):
= Length * Width * Height
= (22 – 2x) * (15 – 2x) * x
= 330x – 74x2 + 4x3
Essentially differentiating is done by taking an equation:
xn and applying the principle nx(n-1)
Therefore if: y = 330x – 74x2 + 4x3 =>dy= 330 – 148x + 12x2
dx
resulting in a quadratic equation, which can be solved by:
x = -b + or - √b2 – 4ac hence x = - (-148) + or - √ (148)2 – 4 (330)(12)
2a 2(12)
x = 148 + or - √ 21904 - 15840
24
x = 9.411cm (4s.f) and 2.922cm (4s.f)
So far my prediction is correct but I think it is necessary to investigate even more values for the length of the rectangle.
21cm
x cm
15cm (15cm – 2x)
x cm
(21cm – 2x)
Question 2d) Similarly for a rectangle with a length of 21cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):
= Length * Width * Height
= (21 – 2x) * (15 – 2x) * x
= 315x – 72x2 + 4x3
Essentially differentiating is done by taking an equation:
xn and applying the principle nx(n-1)
Therefore if: y = 315x – 72x2 + 4x3 =>dy= 315 – 144x + 12x2
dx
resulting in a quadratic equation, which can be solved by:
x = -b + or - √b2 – 4ac hence x = - (-144) + or - √ (144)2 – 4 (315)(12)
2a 2(12)
x = 144 + or - √ 20736 - 15120
24
x = 9.122(4s.f) and 2.878cm (4s.f)
20cm
x cm
15cm (15cm – 2x)
x cm
(20cm – 2x)
Question 2e) Similarly for a rectangle with a length of 20cm, a width of 15cm and a cutout size of x cm, a box can be formed of volume (y):
= Length * Width * Height
= (20 – 2x) * (15 – 2x) * x
= 300x – 70x2 + 4x3
Essentially differentiating is done by taking an equation:
xn and applying the principle nx(n-1)
Therefore if: y = 300x – 70x2 + 4x3 =>dy= 300 – 140x + 12x2
dx
resulting in a quadratic equation, which can be solved by:
x = -b + or - √b2 – 4ac hence x = - (-140) + or - √ (140)2 – 4 (300)(12)
2a 2(12)
x = 140 + or - √ 19600 - 14400
24
x = 8.838(4s.f) and 2.829cm (4s.f)
I have investigated five different rectangles, the results of which can be seen on the spreadsheet below:
Final Conclusion:
As you can see from the graph and the spreadsheet, my final prediction was right. As the length and width values get closer together, the optimum cutout size gets smaller because the volume size decreases.
I feel that my results have been very accurate because not only did I solve various formulae using differentiation but I also checked the results by constructing charts and graphs which showed me the optimum values. This has led to valid conclusions especially for the square Max Box where I achieved a general formula. I know that my other conclusion is valid because I have tested out the theory by doing lots of examples.
I feel that my results were also relevant because once I had found out that the cutout size affects the volume of the box, I was able to continue further in order to see whether a pattern emerged when investigating how the size of the length affected the cutout size when the width of the box remained constantly at 15cm. Unfortunately, I was unable to find a general formula for the rectangle Max Boxes.
