2/6L=2/27 X 18³ = 432cm³
This confirms the first result. Now I will use trial and error to again confirm the maximum value:
Results in the table so far agree that 432cm³ is the maximum volume.
Investigation 2
I will now use a tray using Length=15cm.
This can also be used to prove the shopkeeper is correct.
(15-2x)² = 4(15-2x)
÷(15-2x) ÷(L-2x)
15-2x = 4x
15 = 6x
X = 2.5
V = (15-2x)(15-2x)
V = (225-30x+4x²)x
V = 225x-60x²+4x³
V = 250cm³
This formula shows that when the height of the sides equal 2.5, the trays volume will be at its maximum.
Investigation 3
To further my investigation on the 18x18cm tray I will use differentiation. Differentiation finds the maximum point. By differentiating a function we get the value of any gradient at any particular point. This is called the derivative.
Y = Volume
x = Height
x = 18
y = x(18-2x)²
= x(324-72x+4x²)
= 324-72²+4x³
Dy/dx = 324-144x+12x²
Maximum = 324-144x+12x²=0
(18-2x)²-72x+12x²=0
Area of base= (18-2x)² = 72x-8x²
Area of side= 4x(18-2x)
The above shows that the volume (y in this case) is at a maximum when the area of the base and the area of the side are equal. This shows the shopkeepers claim to be correct. I will continue and find out te maximum volume using the length (w):
y = x(w-2x)²
= x(w²-4wx+4x²)
= xw²-4wx²+4x³
Dy/dx = w²-8wx+4x²
Meaning: w²-8wx+12x²=0
X = 12x²-8wx+w²=0
X = 8w±64w²-48w²
24
= 8w±16w²
24
= 8w±4w
24
= 4w/24 =w/6
12w/24 =w/2 (Cant be w/2 as base would be too small)
I will test that theory by substituting W as been 8.
(8-2x)² = 4x(8-2x)
64-32x+4x² = 32x-8x²
12x²-64x+64 = 0
3x²-16x+16 = 0
(3x-4)(x-4) = 0
X = 4/3
This again proves the shopkeepers claim to be correct! I conclude from this investigation that the shopkeepers claim is correct but I will further my investigation by trying the shopkeepers theory on rectangular trays.
Investigation 4
I will now try the shopkeeper’s theory with rectangular trays. ‘When the area of the base is equal to the area of all four sides the volume of the tray will be at a maximum’.
Area of base = LW
Area of sides = 2(Lx)+2(Wx)
It proves to be more difficult to find a rectangle where the area of both the sides and base are equal. The closest match is:
Height: 2cm
Length: 11cm
Width: 8cm
I believe that this method of finding an exact match for the base area and side area will take too long and wouldn’t be the best method of proving the shopkeepers claim for a rectangle.
If I assume that the shopkeeper is correct and place down that formula I can work from that in order to find a maximum.
(18-2x_(12-2x) = 2x(18-2x)+2x(12-2x)
216-60x+4x² = 36x-4x²+24x-4x²
12x²-120x+216 = 0
X²-10x+18 = 0
A=1
B=-10
C=18
X=
Not possible Where decimal search would end up
Base area=
Area of sides=
Volume=
After investigation number one I found that the shopkeepers claim to be correct using a square based try and substituting lengths into a formula. I further checked the shopkeepers claim by using a trial and error table. This also showed the theory to be correct.
I also tried the same investigation method in investigation two but this time using a 15x15 square. This also showed the claim to be correct.
I then tried differentiation on a 18x18 sqaure. This showed the volume (y) is at a maximum when the area of the base and the area of the sides are equal. Therefore confirming the claim to be correct.
I conclude that the shopkeepers claim is correct for both square and rectangular trays after carrying out my investigations using varying mathematical means.
Robert Davison