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Chapter 4: Problem 31

A cylindrical can, open at the top, is to hold \(500 \mathrm{~cm}^{3}\) of liquid. Find the height and radius that minimize the amount of material needed to manufacture the can.

Short Answer

Expert verified
Radius is \(\sqrt[3]{\frac{1000}{2\pi}}\) and height is \(\frac{500}{\pi (\sqrt[3]{\frac{1000}{2\pi}})^2}\).

Step by step solution

01

Understand the Problem

The problem is to find the dimensions of a cylindrical can with no top that will hold a volume of 500 cubic centimeters and minimize the material needed for its construction. The can's volume can be expressed as \( V = \pi r^2 h = 500 \), and its surface area (material needed) is the area of the side plus the bottom, which is \( S = \pi r^2 + 2\pi rh \).
02

Express Height in Terms of Radius

Since we know the volume \(V = \pi r^2 h = 500\), we can express the height \(h\) in terms of the radius \(r\):\[ h = \frac{500}{\pi r^2}. \]
03

Substitute Height into Surface Area Expression

Substitute the expression for \(h\) from the previous step into the surface area formula. The new expression for surface area becomes: \[ S = \pi r^2 + 2\pi r \left(\frac{500}{\pi r^2}\right) = \pi r^2 + \frac{1000}{r}. \]
04

Find the Derivative of the Surface Area

To minimize the surface area, take the derivative of \(S\) with respect to \(r\) and set it to zero: \[ S' = 2\pi r - \frac{1000}{r^2}. \] Then solve \( S' = 0 \) for \(r\).
05

Solve the Derivative Equation

Set the derivative equal to zero to find \( r \): \[ 2\pi r - \frac{1000}{r^2} = 0. \] Solve for \(r\) to get: \[ 2\pi r^3 = 1000, \] \[ r^3 = \frac{1000}{2\pi}, \] \[ r = \sqrt[3]{\frac{1000}{2\pi}}. \]
06

Calculate the Height

Once we have the radius \(r\), use the equation \( h = \frac{500}{\pi r^2} \) to find the height \(h\). After calculating, you find \(h = \frac{500}{\pi (\sqrt[3]{\frac{1000}{2\pi}})^2} \).
07

Verify the Solution

Use the second derivative test on \( S'' = 2\pi + \frac{2000}{r^3} \) to ensure it is positive, confirming a minimum at the calculated radius. Also, verify that the obtained \(r\) and \(h\) satisfy the volume condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cylindrical Can Design
Designing a cylindrical can involves determining its radius and height to achieve certain requirements while optimizing another aspect, such as minimizing the material used. Here, we focus on a can open at the top, which means it has a circular bottom and a cylindrical side.
The volume of this cylindrical can, which is the space it occupies, should be constant at 500 cubic centimeters. That gives us our first equation:
  • Volume equation: \( V = \pi r^2 h = 500 \)
where \( r \) is the radius and \( h \) is the height of the cylinder. In this design problem, apart from maintaining this volume, we aim to minimize the surface area because that correlates with the material required to construct the can.
Surface Area Minimization
Minimizing surface area is crucial because it directly affects the amount and cost of the material used in manufacturing the can. In our scenario, the can's surface area is the sum of the area of the circular base and the lateral side without the top.
The formula for surface area becomes:
  • Surface area equation: \( S = \pi r^2 + 2\pi r h \)
Initially, we cannot minimize the surface area directly since it has two variables, \( r \) and \( h \). To overcome this, express \( h \) in terms of \( r \) using the volume equation. Doing so simplifies the problem into a single-variable function. This new expression simplifies to:
  • Substituted surface area: \( S = \pi r^2 + \frac{1000}{r} \)
By achieving a single-variable function, we can apply calculus techniques to find the optimum dimensions that minimize the surface area.
Derivative Test for Optimization
To identify the optimal radius that minimizes the surface area, we need to use calculus, specifically, the derivative test.
First, calculate the derivative of the surface area function with respect to \( r \), which involves differentiating \( S = \pi r^2 + \frac{1000}{r} \). This yields:
  • Derivative: \( S' = 2\pi r - \frac{1000}{r^2} \)
By setting this derivative to zero, we find the critical points, i.e., values of \( r \) that could represent the minimum surface area:
  • Equation set to zero: \( 2\pi r - \frac{1000}{r^2} = 0 \)
  • Solving gives: \( r = \sqrt[3]{\frac{1000}{2\pi}} \)
After finding \( r \), solve for \( h \) using the equation \( h = \frac{500}{\pi r^2} \). Finally, confirm that these values indeed minimize the surface area using the second derivative test. If \( S'' > 0 \), the critical point is a minimum. This ensures the solution is not only stationary but optimal for minimum material usage.

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