Volume = h(18-2h)²
= h(18-2h)(18-2h)
= h(324-72h+4h²)
= 324h-72h²+4h³
As shown above, the formula expands to 324h-72h²+4h³, which is a cubic function so when graphed, it will look like:
This graph as a maximum and minimum value and as I have shown opposite, when the graph is at its maximum and minimum, the gradient = 0. So, in order to find the maximum (and minimum) value, the gradient function must = 0. First of all, I must find a gradient function for this graph. To do this, I will differentiate the function for this graph:
Volume = 324h-72h²+4h³
dv = 324-144h+12h²
dh
To find the maximum volume, let gradient = 0:
dv = 324-144h+12h²
dh
- = 324-144h+12h²
- = 12(27-12h+h²)
0 = 12(h-3)(h-9)
h = 3, h = 9
The maximum value of h must = 3, because as the length of the side = 18cm, if h = 9, then the length of the side = 0, and the volume would also = 0.
I then substituted h into the original formula (not differentiated) to find the maximum volume: h(18-2h)²
= 3(18-2x3)²
= 3x144
= 432 cm³
I will now prove the shopkeeper’s theory by substituting h into the formulae representing the area of the sides and the area of the base. If the area of the sides = the area of the base when I substitute h = 3 into the formulae, then the shopkeeper’s theory is correct for an 18 x 18 square.
AB = (18-2h)²
AS = h(18-2h)4
When h = 3, AB = AS
(18-2h)² = h(18-2h)4
(18-2x3)² = 3(18-2x3)4
(18-6)² = 3x12x4
- cm² = 144cm²
Conclusion
For an 18 x 18 square, area of base does = area of side when the volume is maximum, so the shopkeeper’s theory is correct.
I will investigate further whether his theory is correct for all squares. This can be done by coming up with a general formula for AB, AS and volume.
Let the length of the card = x instead of 18.
AB = (x-2h)²
AS = h(x-2h)4
Volume = h(x-2h)²
As before, to maximise the volume, I will differentiate the function for volume:
Volume = h(x-2h)²
= h(x²-4hx+4h²)
=hx²-4h²x+4h³
Although there appears to be 2 variables (h and x) I am able to differentiate this formula, as in the formula, x is always a constant number – the number being whatever the length of the side of card will be. This means I can treat x as just a number, and so differentiate the formula:
Volume =hx²-4h²x+4h³
dv = x²-hx+12h²
dh
To find maximum volume, let dv (gradient) = 0
dh
0 = x²-hx+12h²
This is now the formula representing the maximum volume. According to the shopkeeper, AB=AS when the volume is maximum. I know that the maximum volume is when dv = 0, so AB should = AS when dv = 0.
dh dh
I have already found out that dv = x²-hx+12h² from differentiation, so the
dh
AB = AS when 0 = x²-hx+12h²
Maximum volume: 0 = x²-hx+12h²
AB = AS when 0 = x²-hx+12h²
(x-2h)² = 4(x-2h)h
To prove the shopkeeper’s theory correct, the above formula should simplify to be exactly the same as the formula for the maximum volume
(0 = x²-hx+12h²).
(x-2h)(x-2h) = 4(hx-2h²)
x²-4hx+4h² = 4hx-8h²
x²-4hx+4h²-4hx+8h² = 0
x²-8hx+12h² = 0
Conclusion
In conclusion, the formulae are exactly the same, therefore the shopkeeper’s theory is correct.
Investigating Further
I will now investigate the shopkeeper’s theory even further by seeing if it is true for rectangles as well as squares.
Again, as with the squares, I will perform a 3 variable decimal search in Excel. To do this I must come up with algebraic formulae to represent AB, AS and volume.
Let length of rectangle = L
Let width of rectangle = W
The length of the side = L – length of 2 corners (h), therefore
Length of side = L–2h
The width of the side = W – length of 2 corners (h), therefore
Width of side = W – 2h
In order to find the area of the base, I need to times the length of the side by the width.
The area of the base (AB) = (length of side)(width of side)
AB = (L-2h)(W-2h)
In order to find the area of the sides, I need to multiply the length of the side by h. I then need add on width of the side multiplied by h.
The area of the sides (AS) = (length of side)h + (width of side)h
AS = (L-2h)h + (W-2h)h
In order to find the total area of the sides, I then need to multiply (L-2h)h by 2, as there are two identical sides like this in the rectangle, and (W-2h)h by 2 also.
Total AS = 2(L-2h)h + 2(W-2h)h
In order to find the volume of the tray, I need to multiply the length of the side by the width of the side by the height. The height depends on the value of h (increase value of h = increase value of height):
Volume = h(L-2h)(W-2h)
Excel work
In my excel work, I have proven that the shopkeeper’s theory is correct for 4 rectangles with the following measurements:
- L=18, W=16
- L=18, W=15
- L=17, W=15
- L=17, W=14
I have chosen to vary the length and width in this systematic way so that once I have found the value of h when the volume is at it’s maximum for one rectangle, I know within which range I should search for h again for the next rectangle.
I have chosen to only search h to 2 decimal places because:
- if I searched further it would take a long period of time
- the shopkeeper would not be able to measure to any further degree of accuracy than 2 decimal places.
Due to the fact that I only searched for h up to 2 decimal places, AB does not exactly = AS when the volume is maximum. However, if these measurements are rounded to the nearest whole number, then AB would = AS.
Maximising the volume
Expand: h(L-2h)(W-2h)
= h(LW-2hL-2hW+4h²)
= hLW-2h²L-2h²W+4h³
Differentiate : hLW-2h²L-2h²W+4h³
As with the squares, I can differentiate this function although there appears to be 3 variables (L, W and h) but L and W are numbers which stay constant within the function, and so I am able to treat them as just simple numbers.
dv = LW- 4hL-4hw+12h²
dh
Maximise the volume by letting dv = 0
dh
0 = LW-4hL-4hW+12h²
In order to find out whether the shopkeeper’s theory is correct or not, I will see if this gradient function = the area of the base = area of the side.
AB = (L-2h)(W-2h)
AS = 2(L-2h)h + 2(W-2h)h
If the shopkeeper is correct, then AB should = AS, which should equal to the maximum volume.
Therefore: (L-2h)(W-2h) should = 2(L-2h)h + 2(W-2h)h
And this formula should simplify to be the same as: 0 = LW-4hL-4hW+12h²
AB = AS
(L-2h)(W-2h) = 2(L-2h)h + 2(W-2h)h
LW-2hL-2hW+4h² = 2(hL-2h²) + 2(hW-2h²)
LW-2hL-2hW+4h² = 2hL-4h² + 2hW-4h²
LW-2hL-2hW+4h² = 2hL-8h²+2hW
LW-2hL-2hW+4h²-2hL+8h²-2hW = 0
LW-4hL-4hW+12h² = 0
Conclusion
In conclusion, the two formulas are exactly the same, and so the shopkeeper’s theory is correct for all rectangles and all squares – “when the area of the base is the same as the area of the sides, the volume of the tray will be at it’s maximum.”
