1
7
 DIFFRENTIATION 
Candidate Name:
Tracey-Ann Llewellyn
Candidate Number:
1000671698
School:
Manchester High School
Centre Number:
100067
Teacher’s Name:
Mr. N. Palmer
Territory:
Jamaica
Year of Submission:
2022
2
ACKNOWLEDGEMENT .................................................. Ошибка! Закладка не определена.
PROJECT TITLE................................................................ Ошибка! Закладка не определена.
PURPOSE OF PROJECT/ PROBLEM STATEMENT ..... Ошибка! Закладка не определена.
MATHEMATICAL FORMULATION .............................. Ошибка! Закладка не определена.
THE PROBLEM SOLUTION ........................................................................................................ 9
APPLICATION OF SOLUTION ....................................... Ошибка! Закладка не определена.
DISCUSSION OF FINDINGS/ CONCLUSION ......................................................................... 13
3
Many people have assisted me in successfully completing this assessment.
First and foremost, I thank God for blessing me with the determination, mental health, and
knowledge necessary to execute this endeavor successfully. I'd also like to express my gratitude
to Mr. Palmer, my Additional Mathematics instructor, who guided me through this assignment
and taught me a lot. His recommendations and instructions were really helpful in completing my
School-Based Assessment.
Finally, I would like to express my gratitude to my parents and friends, who have provided me
with invaluable advice and suggestions throughout the project's development.
4
An investigation to determine the maximum volume of a cuboid-shaped container.
5
The Manchester Infirmary's proprietor is Mrs. Thompson. She had purchased a 60m 
30m sheet of metal to construct a metal tank to serve as the Infirmary's main water source after
the residents have been receiving insufficient water supply for a few months. This tank was
intended to be built in the building's backyard, where it would be easier and faster to fill with
rain. The tank would have to make the best use of the materials available in order to hold the
most amount of water. Mrs. Thompson seeks guidance from Mr. Attery, a Mathematics teacher,
in order to handle the dilemma. Mr. Attery will attempt to employ calculus concepts to augment
the intended placement area with the materials available, with the help of some of his students.
Parish council building a shelter
Interior measurement
Big enough to hold a certain number of ppl
6
30m
60m
Figure 1.0: Rectangular sheet of metal 30m * 60m
7
X
X
X
X
X
X
Figure 1.1: The rectangular sheet of metal 60m* 30m with four X*X
8
L
Length of the base is
60-2Xm.
2X is subtracted
because X is cut
from both ends of
each edge.
H
The height is
X, which is
the size of
the edge to
be folded
up.
Figure 1.2: The rectangular sheet with side flops folded up to
make an open top tank.
With of the base is 302Xm.
2X is subtracted
because X is cut from
both ends of each
edge.
Hence, the Volume can be found by:
V(X)= (Length) (Width) (Height)
V(X)= (60-2X) (30-2X)(X)
9
I.
Volume (𝜈)= Length (l)*Width (w)*Height (h)
Since 𝜒 is to be subtracted from the corners of the rectangular shaped tank in order for it to
become a container, the volume of the container would become:

ν=(30-2χ)*(60-2χ)*χ
𝜈 = (1800 − 60𝜒 − 120𝜒 + 4𝑥 2 )𝜒
𝜈 = (1800 − 180𝜒 + 4𝑥 2 )𝜒
𝜈 = 1800𝜒 − 180𝑥 2 + 4𝜒 3
𝜈 = 4𝜒 3 − 180𝑥 2 + 1800𝜒
.: y = 4𝜒 3 − 180𝑥 2 + 1800𝜒
II.
Differentiation

𝑑𝑦
𝑑𝑥
= 3(4)𝜒 3−1 – 2(180)𝑥 2−1 + 1(1800)𝜒1−1
= 12𝑥 2 − 360𝜒 + 1800
10
III.
Factorization

Using the Quadratic Formula
𝑥=
𝜒=
𝜒=
𝜒=
−(−360)±√(−360)2 −4(12)(1800)
2(12)
360±√129600−86400
24
360±√43200
24
360+√43200
24
𝜒 = 23.66
OR
𝜒=
−𝑏±√𝑏 2 −4𝑎𝑐
360−√43200
24
𝜒 = 6.34
.: 𝜒 = 23.66 or 𝜒 = 6.34
2𝑎
to factorize, to determine values of 𝜒
11
In order to prove that the solution is a maximum, the second derivative has to be carried out. It is
proven by using the value of the second derivative at a point.
The test states that: if the function f is twice differentiable at a critical point, х i.e. f(x)=0, then:
If ƒˡˡ (x) < 0 then f has a local maximum at x
If ƒˡˡ (x) > 0 then f has a local minimum at x
If ƒˡˡ (x) = 0 then the test is inconclusive
First derivative 12 X2 - 360 X + 1800
Differentiate to find the second derivative
12 X² - 360X + 1800
2(12)X²-ˡ -1(360)X ˡ-ˡ + 1800
The second derivative is Vˡˡ (X) = 24X – 360
= 24X – 360 < 0
X = 6.34
Vˡˡ (6.34) = 24(6.34) – 360
= -205.68
Hence the volume is maximized at 6.34.
12
Maximum volume = V(X) = (30 -2X) (60 – 2X) X
(6.34) = (30 – 2 [6.34]) (60 – 2 [6.34] ) (6.34)
= (30 – 12.68) (60 – 12.68) (6.34)
= (17.32) (47.32) (6.34)
= 5,196.152416 m³
= 5,196.15 m³ (2dp)
Thus using the principle of calculus, the maximum volume of the metal tank from a 60m × 30m
sheet can hold a maximum of 5,196.15 m³ of water.
13
Mrs. Thompson, the Manchester Infirmary's owner, is keen to build a metallic tank to
serve as the facility's main water source, ensuring that the inhabitants have enough water on a
regular basis. She recruited Mr. Attery and his Additional Mathematics students' assistance in
determining the maximum volume of water that can be held in a tank, which will be constructed
from 30m by 60m of metal that has already been acquired. That is, he will need to know the
tank's dimensions.
To find the maximum volume, Mr. Attery and his students chose to employ calculus
principles, specifically differentiation. The volume of the tank that will store the maximum
amount of water will be 5,196.15 m3 (17.32m*47.32m*6.34m) based on the analysis of the data
collected. However, it was discovered that some metal bits were lost when the group's method
was used.
14
Method #1
The first method, which would result in a greater volume, even though inquiring more loss of
material, would be to construct open oval tank.
1. Cut off one fifth of the metal sheet – 30m*12m.
2. Out of this piece, cut an oval of area 125 m2 (i.e., 6m*15m*22/7).
3. Assemble as shown below.
30m
48m
60 m
30m
48 m
6m
30m
15 m
Figure1.1. Dissection of metal sheet to construct an open oval tank
15
30m
V = Area of Base *Height
= 125m2*30m
=3750m3
30m
Figure 1.2 Three-dimensional diagram of the open oval tank.
Therefore, this method would allow for the storage of a greater volume of water; as curved
surfaces have a greater surface area than straight surfaces.