Optimizing Cake Tin Volumes Using Calculus and Formulas

Introduction

In the real world, numerous practical scenarios require us to determine the maximum or minimum values of a function, often referred to as the optimum solution, through a process known as optimization.

Our specific challenge involves a cake tin manufacturer who produces cake tins of varying sizes, ranging from "tiny" to "gigantic." These tins may have either square or rectangular bases, but in all cases, the manufacturer aims to maximize the volume of each cake tin.

Aim

The aim of this task is to investigate the size of a square cut into a piece of tinplate to form an open-top cake tin while optimizing its volume.

We will employ various mathematical calculations involving differential calculus and the quadratic formula to develop a conjecture, which will then be validated with accurate results.

Method

Our investigation is divided into two parts:

Part A: Investigating a Square Tinplate

An open-top cake tin will be created by cutting a square from each corner of a square piece of tinplate with side lengths of l cm. Following the cuts, the sides are folded to form the open-top cake tin. Let x cm represent the side length of the square cuts to be made.

Part A consists of three subsections:

1. Derivation of the Volume Expression
2. Finding x for Different l Values
3. Determining the Maximum Volume

1. Derivation of the Volume Expression

The volume of a square-based tin can be calculated by multiplying the area of the base by the height of the tin, using the formula: ^V(x) = (l x w) (h).

For a cake tin with a 5 cm side length, the volume, ^V(x), is given by:

^V(x) = (l - 2x) (l - 2x) (x)

^V(x) = (5 - 2x) (5 - 2x) (x)

^V(x) = 25x - 10x² - 10x² + 4x³

^V(x) = 25x - 20x² + 4x³

The derivative of ^V(x), denoted as ^V'(x), is calculated as:

^V'(x) = 25(1) - 20(2x) + 4(3x²)

^V'(x) = 25 - 40x + 12x²

We can apply the quadratic formula, x = (-b ± √(b² - 4ac)) / (2a), to determine the exact value of x that maximizes the volume of the cake tin.

Using the values a = 12, b = -40, and c = 25:

x = (40 ± √(〖(-40)〗² - 4(12)(25))) / (2(12))

x = (40 ± √(1600 - 1200)) / 24

x = (40 ± √400) / 24

Now, we calculate x for at least three other values of l cm, the side lengths of the original square tinplate:

x = (40 ± 20) / 24

x = (40 + 20) / 24 = 60/24 = 5/2

OR

x = (40 - 20) / 24 = 20/24 = 5/6

Finally, we find the exact value of x that maximizes the volume of the cake tin:

When x = 0.83,

^V(x) = 25(0.83) - 20(〖0.83)〗² + 4(〖0.83)〗³

^V(x) = 20.75 - 13.78 + 2.29

^V(x) = 9.26 cm³

Substituting x = 0.83 into the equation for the width, W = L = 5 - 2x:

W = L = 5 - 2(0.83) = 3.34 cm

Therefore, the volume of the cake tin is maximized when L = 5 cm and x = 0.83 cm.

Analysis and Conjecture

The following table presents our investigation into three different values of L. It compares the values of x derived from the application of calculus and the quadratic formula, along with the corresponding volume, ^V(x), in cubic centimeters:

L Value (cm)	x Value (cm)	V(x) (cm3)
3	3/6	2
5	5/6	9.26
8	8/6	37.91

From the data presented in the table, it can be observed that L/x is equivalent to 6, which can be rearranged as L = 6x and x = L/6. A conjecture can be formulated based on these observations, applying to all real values of x, as demonstrated by the four values of L investigated above.

This conjecture can be substantiated through algebraic reasoning. For instance, if the length of the cake tin is represented as L - 2x, then:

^V(x) = L x W x H

^V(x) = (l - 2x) (l - 2x) (x)

^V(x) = L² - 4xL + 4x² (x)

^V'(x) = L² - 4(2x)L + 4(3x²)

^V'(x) = L² - 8xL + 12x²

We can then apply the quadratic formula, x = (-b ± √(b² - 4ac)) / (2a), to determine the value of x that maximizes the volume:

x = (8 ± √(〖-8〗² - 4(12)(1))) / (2(12))

x = (8 ± √(64 - 48)) / 24

x = (8 ± √16) / 24

x = (8 ± 4) / 24

x = (8 + 4) / 24 = 12/24 = 1/2

OR

x = (8 - 4) / 24 = 4/24 = 1/6

This equation holds true for all values of l and x up to infinity, indicating the existence of numerous possibilities, factors, and variables that can influence the value of x. To further validate the conjecture, we will investigate the case where L = 25 to ensure the general equation produced remains accurate.

Applying the conjecture assumption to discover x for L = 25 yields x = (25)/6 = 25/6. We can calculate the corresponding volume, ^V(x), as follows:

^V(x) = L x W x H

^V(x) = (25 - 2x) (25 - 2x) (x)

^V(x) = 625 - 100x + 4x² (x)

^V(x) = 625x - 100x² + 4x³

^V'(x) = 625 - 200x + 12x²

We can use the quadratic formula to determine x:

x = (200 ± √(〖-200〗² - 4(12)(625))) / (2(12))

x = (200 ± √(40000 - 30000)) / 24

x = (200 ± √10000) / 24

x = (200 ± 100) / 24

x = (200 + 100) / 24 = 300/24 = 25/2

OR

x = (200 - 100) / 24 = 100/24 = 25/6

Therefore, we have obtained the same results as the conjecture, confirming that x = 25/6.

In Part B, we delve into the examination of a rectangular tinplate. Our objective is to create an open tin cake by cutting a square with side length x cm from each corner. This analysis aims to formulate and validate a conjecture regarding the relationship between x, the square cut size, and the lengths of each side of the rectangle required to maximize the cake tin's volume.

To develop and substantiate our conjecture, we need to provide evidence by investigating various lengths for r and s and determining their relationship with x. For instance, let r = 3 cm and s = 2 cm be our initial investigation:

Investigating the Value of x for r = 3 cm and s = 2 cm

Given: r = 3 – 2x, H = x, s = 2 – 2x

Limitations: r > 0 cm, s > 0 cm, x > 0 cm since r, s, and x are lengths; x < 1 cm if s = 2 - 2x cm, hence, 0 < x < 1.

This relationship can be demonstrated with algebraic calculations:

2 - 2x > 0

2x < 2

x

The volume, ^V(x), is calculated as:

^V(x) = r x s x H

^V(x) = (3 – 2x) (2 – 2x) (x)

^V(x) = x (6 – 10x + 4x²)

^V(x) = 6x – 10x² + 4x³

The derivative, ^V'(x), is calculated as:

^V'(x) = 6(1) – 10(2x) + 4(3x²)

^V'(x) = 6 – 20x + 12x²

We can now apply the quadratic formula to determine the value of x:

x = (20 ± √(〖-20〗² - 4(12)(6))) / (2(12))

x = (20 ± √(400 - 288)) / 24

x = (20 ± √112) / 24

x = (20 ± 4√7) / 24

x = (20 + 4√7) / 24 = (5 + √7) / 24 ≈ 1.27

OR

x = (20 - 4√7) / 24 = (4(5 - √7)) / 24 = (5 - √7) / 6 ≈ 0.39

Substituting x = 0.39 into ^V(x) = 6x – 10x² + 4x³, r = 3 – 2x, and s = 2 – 2x:

^V(x) = 6(0.39) – 10(0.39)² + 4(0.39)³

r = 3 – 2(0.39) = 2.22 cm

s = 2 – 2(0.39) = 1.22 cm

Therefore, the volume of the cake tin is maximized for r = 3 cm and s = 2 cm when x = 0.39 cm.

We can extend our investigation to other values of r and s and compile the findings into the following table, which follows the same methodology:

r Value (cm)	s Value (cm)	x Value (cm)	V(x) (cm3)
3	2	(5 - √7)/6	1.06
5	3	(8 - √19)/6	4.11
8	4	(12 - √48)/6	12.32

Based on the data from this table, we can establish a relationship between r, s, and x. A general equation for all real values of x can be derived. It can be observed that x = ((r + s) - √(r² - rs + s²)) / 6 holds true for all values of r and s in the table that have been investigated. This equation encompasses all values of r, s, and x up to infinity.

We will now explore the case of a rectangle with sides in the ratio of a:b. For our investigation, we will consider a rectangle with a 2:1 ratio, and we will determine the values of x that maximize the cake tin's volume using the quadratic formula. The values chosen for r and s are 12 and 6 cm, respectively:

Investigating the Value of x for r = 12 cm and s = 6 cm (2:1 Ratio)

Given: a = 12 – 2x, H = x, b = 6 – 2x

Limitations: r > 0 cm, s > 0 cm, x > 0 cm since r, s, and x are lengths; x < 3 cm if s = 6 - 2x cm, hence, 0 < x < 3.

We substantiated these limitations through algebraic reasoning:

6 – 2x > 0

2x < 6

x < 3

The volume equation for this scenario is expressed as:

V(x) = a x b x H

= (12 – 2x)(6 – 2x)(x)

= x(72 – 36x + 4x²)

= 72x – 36x² + 4x³

The derivative of this equation with respect to x is:

V’(x) = 72(1) – 36(2x) + 4(3x²)

= 72 – 72x + 12x²

Next, we applied the quadratic formula to find the values of x that maximize the cake tin's volume:

x = (-b ± √(b² - 4ac)) / (2a)

With the coefficients as follows: a = 12, b = -72, and c = 72, we calculated the discriminant (∆) as:

∆ = (-72²) - 4(12)(72)

= 5184 - 3456

= 1728

Solving for x, we get:

x = (72 ± √1728) / (2(12))

= (72 ± √1728) / 24

= (72 ± 24√3) / 24

Which results in:

x = (72 + 24√3) / 24 ≈ 4.73

OR

x = (72 - 24√3) / 24 ≈ (4(18 - 6√3)) / 24 ≈ (18 - 6√7) / 6 ≈ 1.26

Substituting x = 1.26 into the volume equation, with r = 12 - 2x and s = 6 - 2x, we find:

V(x) = 72(1.26) - 36(1.26)² + 4(1.26)³

r = 12 - 2(1.26) = 6 - 2(1.26)

Resulting in:

≈ 90.72 - 45.36 + 5.04 ≈ 12 - 2(1.26) ≈ 6 - 2(1.26)

Thus, the volume of the cake tin is maximized when r = 12 cm and s = 6 cm in a 2:1 ratio, with x approximately equal to 1.26 cm.

The table below apply the same way of investigations into three other values of r and s. This table compares all the values of x that has been discovered from the value of r and s by using calculus and the quadratic formula. From this table a conjecture can be analyse.

r Value (cm)	s Value (cm)	X Value (cm)	V(x) (cm3)
12	6	(18 - 6√3)/6	50.4
9	3	(12 - 3√7)/6	8.53
28	7	(35 - 7√13)/6	150.83

Based on the insights derived from this table, a general equation has been formulated that establishes a relationship between s and x:

x = ((s + sn) - s√(n² - n + 1)) / 6

Conclusion

In conclusion, in part A the relationship between x and the length of the sides of a cake-tin is dependent on the cake tin’s lengths. If both sides are in the same length, then the general equation can be predicted any value for x = L/6. This proved with algebra and by using the quadratic formula. In part B if the sides (r and s) of the rectangular tinplate are different, then the general equation to predict the value of x was discovered to be ((r+s)-√(r^2-rs+ s^2 ))/6 where r and s are the sides of the cake tin. Furthermore in part b, when the sides r and s were in particular ratio such as 2:1, the formula to find x was similar to the previous formula except everything under the square root must be multiply by s, which lead into the general equation x=((s+sn)-s√(n^2-n+1))/6, where s is the shorter side of the cake tin and n is the ratio of the sides. In all the assumption equation, the length of the cake tin sides is always the numerator, while the denominator is 6. This investigation has provided insight into the patterns and relationships between l and x, and r, s and x, and have also allowed the ability to provide evidence of differential calculus knowledge.