Problem 15 - Thomas Calculus in SI Units

Question

You are designing a \(1000 \mathrm{cm}^{3}\) right circular cylindrical can whose manufacture will take waste into account. There is no waste in cutting the aluminum for the side, but the top and bottom of radius \(r\) will be cut from squares that measure \(2 r\) units on a side. The total amount of aluminum used up by the can will therefore be $$A=8 r^{2}+2 \pi r h$$ rather than the \(A=2 \pi r^{2}+2 \pi r h\) in Example \(2 .\) In Example 2 the ratio of \(h\) to \(r\) for the most economical can was 2 to \(1 .\) What is the ratio now?

Step-by-Step Solution

Answer

The ratio \(\frac{h}{r}\) is approximately \(\frac{8}{\pi}\) or 2.54648.

Step 1: Understand the Problem

We are given a cylindrical can with a volume of 1000 cm³. The sides are made out of a continuous material with no waste, but the top and bottom are cut from squares measuring \(2r\) on each side. Our task is to find the ratio of height \(h\) to radius \(r\) to minimize the amount of material used, expressed by the equation \(A=8r^2 + 2\pi rh\).

Step 2: Write the Volume Equation

The volume \(V\) of a cylinder is given by \(V = \pi r^2 h\). We know the volume is 1000 cm³, so \(\pi r^2 h = 1000\). We can use this equation to express \(h\) in terms of \(r\).

Step 3: Solve for Height in Terms of Radius

Rearrange the volume equation to solve for \(h\):\[ h = \frac{1000}{\pi r^2} \]

Step 4: Substitute for Height in Material Used Equation

Substitute \(h = \frac{1000}{\pi r^2}\) into the material used equation \(A = 8r^2 + 2\pi rh\):\[ A = 8r^2 + 2\pi r\left( \frac{1000}{\pi r^2} \right) \]Simplify to obtain:\[ A = 8r^2 + \frac{2000}{r} \]

Step 5: Find the Minimum Material Used

To minimize the material used, differentiate \(A\) with respect to \(r\) and set the derivative equal to zero. The derivative is:\[ \frac{dA}{dr} = 16r - \frac{2000}{r^2} \]Set this equal to zero:\[ 16r = \frac{2000}{r^2} \]

Step 6: Solve for Radius

Rearrange and solve for \(r\):Multiply both sides by \(r^2\):\[ 16r^3 = 2000 \]Divide by 16:\[ r^3 = 125 \]Taking the cube root:\[ r = 5 \]

Step 7: Find Height Using Radius

Substitute \(r = 5\) into \(h = \frac{1000}{\pi r^2}\):\[ h = \frac{1000}{\pi (5)^2} = \frac{1000}{25\pi} = \frac{40}{\pi} \]

Step 8: Calculate the Ratio of Height to Radius

Now that we have \(r = 5\) and \(h = \frac{40}{\pi}\), find the ratio \(\frac{h}{r}\):\[ \frac{h}{r} = \frac{\frac{40}{\pi}}{5} = \frac{8}{\pi} \]

Step 9: Interpret the Ratio

The ratio \(\frac{h}{r} = \frac{8}{\pi}\) means that the height is about \( \frac{8}{\pi} \) times the radius, which is approximately \(2.54648\), indicating that the height is longer than the radius in this optimal design with wastage taken into account.

Key Concepts

Cylindrical Geometry

Cylindrical geometry is the study of shapes that resemble a cylinder. The basic elements of a cylinder include the base radius \( r \), height \( h \), and lateral surface. A right circular cylinder is the most common type, featuring circular bases and a perpendicular side.

In this context, we explore a cylindrical can whose volume must remain constant at \(1000 \mathrm{cm}^{3}\). Despite the simple shape, various challenges arise when accounting for materials, especially when considering waste. Understanding the geometric properties of a cylinder helps in solving optimization problems, such as minimizing material use. This involves balancing the base area and side height while meeting volume constraints.

A cylinder's volume \( V \) can be calculated using the formula \( V = \pi r^2 h \). This helps us express the height \( h \) as a function of the radius \( r \). By understanding these relationships, we can derive important equations necessary for optimizing designs for efficiency and material usage.

Derivative Application

In calculus, derivatives are tools used to determine how a function changes as its input changes. This particular problem makes the derivative crucial in identifying the optimal dimensions that minimize material usage.

By taking the derivative of the material use equation with respect to the radius \( r \), we identify specific points where the material used, \( A = 8r^2 + \frac{2000}{r} \), reaches a minimum. The process involves setting the derivative to zero. This step finds the next critical point, where the rate of change switches from decreasing to increasing or vice versa.

For this exercise:

• First, we differentiate \( A \) with respect to \( r \).
• The derivative \( \frac{dA}{dr} = 16r - \frac{2000}{r^2} \) helps find critical points.
• Setting \( 16r = \frac{2000}{r^2} \) and solving reveals the perfect radius for minimizing material usage.

Calculus aids in managing these complex relationships, simplified by applying derivative concepts.

Material Minimization

In design and manufacturing, material minimization is crucial for cost efficiency and sustainability. This can involves not only maintaining necessary dimensions but also managing resource waste.

In this specific scenario, while the sides of the can do not produce waste, the top and bottom do, as they are cut from larger squares. This necessitates revising the calculation of the total material used: \( A = 8r^2 + 2\pi rh \). The former equation reflecting no waste was \( A = 2\pi r^2 + 2\pi rh \).

These changes lead to a different optimal ratio of height \( h \) to radius \( r \). Deriving this new ratio, \( \frac{8}{\pi} \), considers material efficiency more comprehensively by accounting for waste. This means the height, at optimal conditions, is approximately 2.546 times the radius, slightly altering the ratio compared to a waste-free scenario as stated in Example 2.

After solving, the optimal dimensions show the substantial impact of even minimal waste considerations. Understanding these factors is vital for more sustainable, cost-effective designs.