Stage 1 methods
Introduction to Differential Calculus-Cake Tin
Chloe Zhou
Introduction
Calculus is one of the foremost branches of mathematics which involves the study of rates of
change. It was first invented in the mid- 17th century by Sir Isaac Newton. He created calculus during
his studies with gravity. Calculus acts as a dominant tool in Physics, such as motion, electricity, heat,
and light. It has also been widely applied in real life scenario, such as geography, computer vision,
artificial intelligence, robotics, as well as to calculate the rates of radioactive decay in chemistry and
even to predict birth and death rates. One of the most common applications of calculus in practical
circumstances is optimisation, which is the largest value or the smallest value that a function can
take.
A cake tin manufacture creating cake tins that ranging in size from “tiny” to “gigantic”. Some tins will
be square based others will be rectangular based. The demand is to maximise the volume of each
cake tin. This investigation is an example of optimisation, it will explore what value of side length 𝑥
of these four squares will maximise the capacity of open-top cake tins created from a tin plate. This
investigation is divided into two parts, Part A is the mainly focus on the determining the size of the
square base tins, while part B focuses on the rectangular tins. This will be determined through the
concepts of various mathematical calculations involving geometry, algebra, and the second
derivative function . A conjecture will be made and potentially proven through the applications of
mathematical approaches such as calculus and quadratic formula. The limitation will also be
discussed in this investigation.
Mathematical procedure
For the length of the square cut from Part A, it will be determined by differentiated function the
original function. Therefore, the cubic function will be differentiated into a quadratic function, the
roots of the quadratic function is either the local maximum, local minimum or the inflection point of
the cubic function. Since the equation has a degree of 2, two answers will be obtained. And the
derivative of the cubic will be differentiated once more, the second derivative will identify the
points. In further investigation, three different values of length 𝑙 of the original square tin plate will
be chosen in order to compare the results with the initial. A table with the results will be illustrated
and a conjecture of the exact of each square tin plate will be established. In Part B, the sides of the
rectangle are divided into a ratio of r:s. Two other ratios will be chosen so as to compare the result
with the initial ratio 2:1. Same as Part A, four squares with side length 𝑥 will be cut from each end
and the 𝑥 that maximise the volume will be solved through quadratic formula. The concept of the
second derivative function will be utilized to distinguish whether it is a local maximum, local
minimum or the point of inflection and support the proposition. Another table with the answers
obtained in Part B will be listed and a second conjecture will be established. Eventually, a
proposition will be proven by using principles of calculus and the quadratic formula for both of the
conjecture concluded from Part A and B.
Context of the problem
People benefit from the application of optimisation, there are practical scenarios to minimize costs
in the production of a good or service and maximize profits. This is the optimum solution, and the
process is called optimisation. One of the examples will be explored in this investigation. Cakes come
in a variety of shapes and sizes in real life. As a result, various cakes must be baked in various cake
tins which come in square or rectangular shapes with sizes ranging from "small" to "giant." The
demand for optimising the volume of cake tins to have the maximum volume for each of the cake
tins by utilising the least material, so as to increases the economy benefits. Thus, the manufacturer
, maximises the profit. A square from each corner of a piece of tinplate is cut, then the edges of the
tinplate are folded to produce an open top cake tin.
Results
Part A
In part A, a square base cake tin will be construct by cutting out squares of length 𝑥 𝑐𝑚 from each
corner of a square tinplate with side length 𝑙 𝑐𝑚(Figure 1). The squares are cut to fold a cake
tin(Figure 2). Tin plate of length 6cm, 7cm and 8cm will also be applied in order to support the
relationship between 𝑙 and 𝑥. Lastly a conjecture based on the finding will be presented.
𝑙 𝑐𝑚
𝑥 𝑐𝑚
(Figure 2, an illustration of cake tin)
𝑙 𝑐𝑚
‘ 1, a diagram of a square tinplate with side length 𝑙 𝑐𝑚)
(Figure
The volume of a squared-base prism 𝑉(𝑥) = ℎ𝑒𝑖𝑔ℎ𝑡 × 𝑤𝑖𝑑𝑡ℎ × 𝑙𝑒𝑛𝑔𝑡ℎ since the cut-out corners
are 𝑥 𝑐𝑚 each, then the length of each side is (𝑙 − 2𝑥) 𝑐𝑚. The squares are 𝑥 𝑐𝑚 𝑏𝑦 𝑥 𝑐𝑚, therefore
the height is 𝑥 cm. Hence 𝑉(𝑥) = (𝑙 − 2𝑥)(𝑙 − 2𝑥)𝑥 𝑐𝑚3 . This is the formula used the volume of
the cake tin box. The given length of each side of the tinplate is 5cm,
Therefore, 𝑉(𝑥) = (5 − 2𝑥)(5 − 2𝑥)𝑥
𝑉(𝑥) = 25𝑥 − 20𝑥 2 + 4𝑥 3
The length of 𝑥 that maximise the volume is the local maximum of the cubic function above. The
local maximum can be determined by the roots of the derivative function.
𝑉 ′ (𝑥) = 1 × 25𝑥 0 − 2 × 20𝑥 2−1 + 3 × 43−1
∴𝑉 ′ (𝑥) = 25 − 40𝑥 + 12𝑥 2
−𝑏±√𝑏2 −4𝑎𝑐
The function 𝑉’(𝑥) will be substitute into the quadratic formula 𝑥 = 2𝑎
to obtain an exact
value, where 𝑎 = 12, 𝑏 = −40 and 𝑐 = 25.
25 − 40𝑥 + 12𝑥 2 = 0
−(−40)±√(−40)2 −4×12×25
𝑥=
2×12
5 5
Therefore, the two 𝑥 values are either 6 and 2 .
The length 𝑙 must be greater than zero:
5 − 2𝑥 > 0
−2𝑥 > −5
5
∴𝑥 < 2
5
∴𝑥 ≠ 2
5
∴the length will between 0 𝑎𝑛𝑑 2
5
The length 6 satisfies the proposition. But another method will be
Introduction to Differential Calculus-Cake Tin
Chloe Zhou
Introduction
Calculus is one of the foremost branches of mathematics which involves the study of rates of
change. It was first invented in the mid- 17th century by Sir Isaac Newton. He created calculus during
his studies with gravity. Calculus acts as a dominant tool in Physics, such as motion, electricity, heat,
and light. It has also been widely applied in real life scenario, such as geography, computer vision,
artificial intelligence, robotics, as well as to calculate the rates of radioactive decay in chemistry and
even to predict birth and death rates. One of the most common applications of calculus in practical
circumstances is optimisation, which is the largest value or the smallest value that a function can
take.
A cake tin manufacture creating cake tins that ranging in size from “tiny” to “gigantic”. Some tins will
be square based others will be rectangular based. The demand is to maximise the volume of each
cake tin. This investigation is an example of optimisation, it will explore what value of side length 𝑥
of these four squares will maximise the capacity of open-top cake tins created from a tin plate. This
investigation is divided into two parts, Part A is the mainly focus on the determining the size of the
square base tins, while part B focuses on the rectangular tins. This will be determined through the
concepts of various mathematical calculations involving geometry, algebra, and the second
derivative function . A conjecture will be made and potentially proven through the applications of
mathematical approaches such as calculus and quadratic formula. The limitation will also be
discussed in this investigation.
Mathematical procedure
For the length of the square cut from Part A, it will be determined by differentiated function the
original function. Therefore, the cubic function will be differentiated into a quadratic function, the
roots of the quadratic function is either the local maximum, local minimum or the inflection point of
the cubic function. Since the equation has a degree of 2, two answers will be obtained. And the
derivative of the cubic will be differentiated once more, the second derivative will identify the
points. In further investigation, three different values of length 𝑙 of the original square tin plate will
be chosen in order to compare the results with the initial. A table with the results will be illustrated
and a conjecture of the exact of each square tin plate will be established. In Part B, the sides of the
rectangle are divided into a ratio of r:s. Two other ratios will be chosen so as to compare the result
with the initial ratio 2:1. Same as Part A, four squares with side length 𝑥 will be cut from each end
and the 𝑥 that maximise the volume will be solved through quadratic formula. The concept of the
second derivative function will be utilized to distinguish whether it is a local maximum, local
minimum or the point of inflection and support the proposition. Another table with the answers
obtained in Part B will be listed and a second conjecture will be established. Eventually, a
proposition will be proven by using principles of calculus and the quadratic formula for both of the
conjecture concluded from Part A and B.
Context of the problem
People benefit from the application of optimisation, there are practical scenarios to minimize costs
in the production of a good or service and maximize profits. This is the optimum solution, and the
process is called optimisation. One of the examples will be explored in this investigation. Cakes come
in a variety of shapes and sizes in real life. As a result, various cakes must be baked in various cake
tins which come in square or rectangular shapes with sizes ranging from "small" to "giant." The
demand for optimising the volume of cake tins to have the maximum volume for each of the cake
tins by utilising the least material, so as to increases the economy benefits. Thus, the manufacturer
, maximises the profit. A square from each corner of a piece of tinplate is cut, then the edges of the
tinplate are folded to produce an open top cake tin.
Results
Part A
In part A, a square base cake tin will be construct by cutting out squares of length 𝑥 𝑐𝑚 from each
corner of a square tinplate with side length 𝑙 𝑐𝑚(Figure 1). The squares are cut to fold a cake
tin(Figure 2). Tin plate of length 6cm, 7cm and 8cm will also be applied in order to support the
relationship between 𝑙 and 𝑥. Lastly a conjecture based on the finding will be presented.
𝑙 𝑐𝑚
𝑥 𝑐𝑚
(Figure 2, an illustration of cake tin)
𝑙 𝑐𝑚
‘ 1, a diagram of a square tinplate with side length 𝑙 𝑐𝑚)
(Figure
The volume of a squared-base prism 𝑉(𝑥) = ℎ𝑒𝑖𝑔ℎ𝑡 × 𝑤𝑖𝑑𝑡ℎ × 𝑙𝑒𝑛𝑔𝑡ℎ since the cut-out corners
are 𝑥 𝑐𝑚 each, then the length of each side is (𝑙 − 2𝑥) 𝑐𝑚. The squares are 𝑥 𝑐𝑚 𝑏𝑦 𝑥 𝑐𝑚, therefore
the height is 𝑥 cm. Hence 𝑉(𝑥) = (𝑙 − 2𝑥)(𝑙 − 2𝑥)𝑥 𝑐𝑚3 . This is the formula used the volume of
the cake tin box. The given length of each side of the tinplate is 5cm,
Therefore, 𝑉(𝑥) = (5 − 2𝑥)(5 − 2𝑥)𝑥
𝑉(𝑥) = 25𝑥 − 20𝑥 2 + 4𝑥 3
The length of 𝑥 that maximise the volume is the local maximum of the cubic function above. The
local maximum can be determined by the roots of the derivative function.
𝑉 ′ (𝑥) = 1 × 25𝑥 0 − 2 × 20𝑥 2−1 + 3 × 43−1
∴𝑉 ′ (𝑥) = 25 − 40𝑥 + 12𝑥 2
−𝑏±√𝑏2 −4𝑎𝑐
The function 𝑉’(𝑥) will be substitute into the quadratic formula 𝑥 = 2𝑎
to obtain an exact
value, where 𝑎 = 12, 𝑏 = −40 and 𝑐 = 25.
25 − 40𝑥 + 12𝑥 2 = 0
−(−40)±√(−40)2 −4×12×25
𝑥=
2×12
5 5
Therefore, the two 𝑥 values are either 6 and 2 .
The length 𝑙 must be greater than zero:
5 − 2𝑥 > 0
−2𝑥 > −5
5
∴𝑥 < 2
5
∴𝑥 ≠ 2
5
∴the length will between 0 𝑎𝑛𝑑 2
5
The length 6 satisfies the proposition. But another method will be
