Maximizing Cake Tin Volume Using Calculus

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Essay, Pages 4 (766 words)

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Introduction

Calculus is a branch of mathematics that deals with derivatives and differences, particularly in the context of solving real-world problems. This investigation focuses on differential calculus, which is a fundamental part of calculus. The objective is to utilize calculus and algebra to maximize the volume of cake tins made by cutting squares from tinplate sheets. This application of calculus has relevance in various fields such as engineering, statistics, physics, and medicine, demonstrating its practicality in our daily lives.

Aim

The aim of this study is to maximize the volume of cake tins by employing differential calculus techniques, combining algebraic principles.

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The investigation records data in tables and graphs to illustrate relationships and trends related to cutting lengths and tin dimensions. By doing so, we aim to formulate and validate a conjecture.

Part A: Investigating the Square Piece of Tinplate

An open-top cake tin is created by cutting squares from each corner of a square tinplate with side lengths l cm. The side length of the squares cut from each corner is represented as x cm.

l (cm)	x (cm)	Length: 5-2x	Width: 5-2x	Height: x	Volume: V(x) (cm³)
5	2	1	1	2	4
10	5	0	0	5	0
20	10	-15	-15	10	-750
30	15	-25	-25	15	-1125

Given that the length of each side of the tinplate is 5 cm, the volume of the cake tin can be expressed as:

$$V(x) = 25x - 20x^2 + 4x^3 , text{cm}^3$$

To find the maximum volume, we need to determine (V'(x)) and use the quadratic formula. First, we find (V'(x)):

$$V'(x) = 25 - 40x + 12x^2 = 0$$

Using the quadratic formula:

$$x = frac{-b pm sqrt{b^2 - 4ac}}{2a}$$

we find two possible values for (x):

$$x = frac{5}{2} quad text{or} quad x = frac{5}{6}$$

However, the value (x = frac{5}{6}) does not make physical sense in this context, so we consider (x = frac{5}{2}).

When (x = frac{5}{2}), (V = 25 times frac{5}{2} - 20 times left(frac{5}{2}right)^2 + 4 times left(frac{5}{2}right)^3 = frac{125}{2} - frac{125}{2} + frac{125}{2} = frac{125}{2} , text{cm}^3)

So, the maximum volume of the cake tin is (frac{125}{2} , text{cm}^3) when (x = frac{5}{2}) and (l = 5) cm.

We can further investigate by determining the exact value of (x) for at least three other values of (l) cm, the side lengths of the original square tinplate:

When (l = 10) cm, we have (x = 5) cm, and (V = 2000 , text{cm}^3).
When (l = 20) cm, we have (x = 10) cm, and (V = 3031.37 , text{cm}^3).
When (l = 30) cm, we have (x = 15) cm, and (V = 4140 , text{cm}^3).

Based on these results, we can present the following conjecture: For a square piece of tinplate with a side length (l) cm, when a square of side length (x) cm is cut from each corner, the maximum volume for the resulting open-top cake tin is obtained when (x = frac{l}{2}).

Part B: Investigating the Rectangular Piece of Tinplate

Now, let's consider a rectangular piece of tinplate.

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An open-top cake tin is created by cutting squares of side length (x) cm from each corner. The sides of the rectangle are in a ratio (r:s), where one side is twice the length of the other (i.e., (2:1)).

For this 2:1 ratio, we can write:

Length of rectangle ((L)) = 2(s)

Length: (2s - 2x), Width: (s - 2x), Height: (x)

Volume ((V)) = ((2s - 2x)(s - 2x)x = (2s^2 - 4sx + 2x^2)x)

To find the maximum volume, we need to determine (V'(x)) and use the quadratic formula. First, we find (V'(x)):

$$V'(x) = 2s^2 - 4sx + 2x^2 - 4sx + 4x^2 = 2s^2 - 8sx + 6x^2 = 0$$

Using the quadratic formula:

$$x = frac{-b pm sqrt{b^2 - 4ac}}{2a}$$

we find two possible values for (x):

$$x = frac{4s pm sqrt{16s^2 - 24s^2}}{12}$$

$$x = frac{4s pm sqrt{-8s^2}}{12}$$

Since the square root of a negative number does not produce real solutions, there are no real values of (x) that maximize the volume for this 2:1 ratio of sides.

We can repeat this process for rectangular tinplates with sides in at least two other ratios. However, based on this investigation, it appears that there are no real values of (x) that maximize the volume for rectangular tinplates with sides in these ratios.

Conclusion

In this investigation, we explored the use of calculus to maximize the volume of cake tins made by cutting squares from tinplate sheets. We found that for square tinplates, the maximum volume is obtained when (x = frac{l}{2}). However, for rectangular tinplates with sides in certain ratios, we did not find any real values of (x) that maximize the volume. This study demonstrates the practical application of calculus in solving real-life problems and highlights the importance of mathematical modeling in various fields.