I shall now try between 2.4 and 2.6.
The maximum value of x is 2.5cm. This shows my prediction was correct. This is that a square box has to be divided by six to find the cut out size in which would give the maximum volume.
To help find the highest value of x to make the largest volume, I will use algebra, with aid from textbook sources. I will need to divide the cut out square by the original length of the box.
L= Length of box
Y= Length of box- 2x
10= 2x+y
x = 10/6 or L/6
I will substitute these into simple equations:
V = (L – 2L/6)(L – 2L/6) L/6
I multiplied the 2 in each bracket by L/6 which gives me:
V = (L – 2L/6)(L- 2L/6)L/6
This makes
V = L (L- 2L/6) – “L/6 (L – 2L/6) L/6
This is:
V = (L² - 2L²/6 – 2L²/6 – 4L²)L/6
I will multiply the lot by L/6
V = L³/6 – 2L³/36 – 2L³/36 + 4L³/216
Now I have put the denominator as 216
V = (4L³ + 36L³ - 12L³ - 12L³)/216
This makes
40L³ - 24L³ / 216
This is simplified as
V = 2L³/ 27
I will now test this using different size squares.
- 10cm by 10cm = 2x10³/27 = 74.074074
- 15cm by 15cm = 2x15³/27 = 250
This proves that the equation is correct.
Also there is another way to find out the highest volume shown below. I will show this formula using a 10 by 10 square.
V = x(10-2x)(10-2x) This is minus 2x because of the lengths of the square cut-outs.
= x(100-20x-20x+4x²)
= 4x³-40x²+100x
Using the gradient function I can work this out. The gradient function is this g = anx n-1.
This makes
12x²-80x+100 = 0
Using ax²+bx+c = 0
x = - b +/- √b²-4ac
2a
x = 80 +/- √-80² - (4x12x100)
2x12
x = 80 +/- √6400 – 4800
24
x = 80 +/- 40
24
If it were minus then this is the result- 1.667
If it were plus then this is the result- 5 This would be impossible as this would make up the length of the square so minus is correct.
I am now going to investigate the open box with a rectangle. I will begin this part of my investigation with a general formula for the volume of the rectangle.
The volume of an open box can be expressed like this:
V = x (L-2x)(W-2x)
This is simplified as:
V = x (4x² + LW – 2Wx – 2Lx)
V = 4x³ + 2Lx² - 2Wx² + LWx
V = x(LW) + 4x³ - x²(2L + 2W)
I can now differentiate When dv/dx = 0
dv/dx = LW + 12x² - 2(2L + 2W)
0 = 12x² - (4L + 4W) + LW
This is now a quadratic equation, we can use this formula:
x = -b ± √-(4L + 4W)² - 4 x 12 x LW
2 x 12
This is simplified as:
x = 4(L + W) ± √-(4L + 4W) – 48LW
24
Multiplying the two brackets together:
x = 4(L + W) ± √16L² + 16W2 – 16LW
24
Factorising:
x = 4(L + W) ± √16 (W² + L² - LW)
24
x = 4(L + W) ± 4√(W² + L² - LW)
24
I can simplify by dividing by the common factor of 4:
x = L + W ± √W² + L² - LW
6
I cannot simplify anymore, but using this I am able to find the value of x which would give me the largest volume of the rectangular box. I shall make predictions to some rectangles and prove them. Firstly I will find the largest volume of the box if the rectangle was 10cm by 15cm.
Firstly my prediction:
x = 10 + 15 ± √15² + 10² - 10x15
6
x = 25 ± √225 + 100 -150
6
x = 25 ± √175
6
x = 25 + 13.22875656 = 6.371459427
6
x = 25 – 13.22875656 = 1.961873907 or 1.962 (3.d.p)
6
x must be 1.962 because otherwise 6.371 (3.d.p) is over twice the length. I shall prove that this formula is correct.
Using the graph I can see that the highest is between 1 and 3.
Into decimal points as it is more accurate between 1and 2.
The highest is between 1.9 and 2 so I will go into 2 decimal places.
From this the highest is 1.96. I shall go into further decimal places to be certain between 1.96 and 1.97.
From this I can tell that 1.962 is the highest value and so my prediction was correct. I shall now try with a rectangle of 20 by 40.
First my prediction:
x = 20 + 40 ± √40² + 20² - 20x40
6
x = 60 ± √1600 + 400 – 800
6
x = 60 ± √1200
6
x = 60 + 34.64101615 = 15.77350269
6
x = 60 – 34.64101615 = 4.226497308 = 4.226 (3.d.p)
6
x must be 4.226 because 15.774 (3.d.p) would be more than the length.
I shall prove this:
This graph and the table shows that the highest volume achieved would be between 4 and 5:
From this table, I can see that the highest volume that will be achieved is between 4.2 and 4.3:
This shows that the answer is between 4.22 and 4.23:
This shows that 4.226cm needs to be cut off to give the highest volume for a box of a 20 by 40cm rectangle. This proves that my formula is correct.
To show that the formula; x = L + W ± √W² + L² - LW
6
Is correct by the formula obtained earlier:
x = -b ± √ b² - 4ac
2a
Also the formula v = x (L-2x)(W-2x)
Using the rectangle 10 by 15:
v = x (10-2x)(15-2x)
v = x (4x² - 50x + 150)
v = 4x³ - 50x² + 150x
Now differentiate:
dv/dx = 4x³ - 50x² + 150x
When dv/dx = 0
0 = 12x² - 100x + 150
Simplified by the common factor of four:
0 = 3x² - 25x + 37.5
This is now ready for the formula:
x = 25 ±√25²-4x3x37.5
2x3
x = 25 ±√175
6
x = 25 + 13.22875656= 6.371459427
6
x = 25 – 13.22875656 = 1.961873907
6
1.962 (3.d.p) is correct as 6.371 (3.d.p) would be too large. This shows that my formula is correct. I shall show this with the 20 by 40 rectangle:
v = (20-2x)(40-2x)
v = (4x²-120x+800)
v = 4x³ - 120x² + 800x
Now differentiate:
dv/dx = 4x³ - 120x² + 800x
When dv/dx = 0
0 = 12x² - 240x + 800
Divide by the common factor of four:
0 = 3x² - 60 + 200
Now I can use the formula:
x = 60 ±√60² - 4x3x200
2x3
x = 60±√3600 – 4x3x200
6
x = 60 ±/1200
6
x = 60 + 34.64101615 = 15.77350269
6
x = 60 - 34.64101615 = 4.226497308
6
4.226 (3.d.p) is correct because 15.774 (3.d.p) is too large to make the box.
These results confirm that the formula is correct:
x = L + W ± √W² + L² - LW
6
This formula shows the size of the cut off which gives the highest possible volume for a rectangular box.