Question

a. Classical problem Find the radius and height of a cylindrical soda can with a volume of \(354 \mathrm{cm}^{3}\) that minimize the surface area. b. Real problem Compare your answer in part (a) to a real soda can, which has a volume of \(354 \mathrm{cm}^{3},\) a radius of \(3.1 \mathrm{cm},\) and a height of \(12.0 \mathrm{cm},\) to conclude that real soda cans do not seem to have an optimal design. Then use the fact that real soda cans have a double thickness in their top and bottom surfaces to find the radius and height that minimizes the surface area of a real can (the surface areas of the top and bottom are now twice their values in part (a)). Are these dimensions closer to the dimensions of a real soda can?

Short answer

If so, how do they compare? Answer: Yes, the optimal design closely matches the dimensions of a real soda can. The optimal dimensions are approximately \(r \approx 3.004\ \mathrm{cm}\) and \(h \approx 12.377\ \mathrm{cm}\), while the real soda can dimensions are radius \(3.1\ \mathrm{cm}\) and height \(12\ \mathrm{cm}\). When adjusting for double-layered top and bottom, the new optimal dimensions are even closer: \(r \approx 3.146\ \mathrm{cm}\) and \(h \approx 11.878\ \mathrm{cm}\).

Step-by-step solution

Establish the relationship between volume and surface area

The volume of a cylinder is defined as \(V = \pi r^2h\), where \(r\) is the radius, and \(h\) is the height. The surface area of a cylinder can be expressed as \(A = 2\pi r^2 + 2\pi rh\) which accounts for the top and bottom circle's area and the lateral area respectively. As the volume is given, we can write the height in terms of the given volume and radius as follows: \(h = \frac{V}{\pi r^2} = \frac{354}{\pi r^2}\).

Substitute the height in terms of the volume and radius into the surface area formula

To find the optimum radius and height, we can substitute the height expression we found in step 1 into surface area formula: \(A(r) = 2\pi r^2 + 2\pi r \cdot \frac{354}{\pi r^2}\). Simplify the expression: \(A(r) = 2\pi r^2 + \frac{708}{r}\).

Differentiate the surface area expression with respect to the radius

To find the minimum surface area point, we need to find the critical points by taking the first derivative of \(A(r)\) with respect to \(r\): \(A'(r) = \frac{d}{dr}(2\pi r^2 + \frac{708}{r}) = 4\pi r - \frac{708}{r^2}\).

Solve for the critical points of the surface area function

To find the critical points, set the derivative equal to zero and solve for \(r\): \(4\pi r - \frac{708}{r^2} = 0 \implies r^3 = \frac{708}{4\pi} \implies r \approx 3.004 \mathrm{cm}\). Now, find the corresponding height using the expression for \(h\) found in step 1: \(h = \frac{354}{\pi r^2} \approx 12.377 \mathrm{cm}\).

Compare the optimal theoretical design to the real soda can dimensions and adjust for the double-layer top and bottom

The optimal dimensions found are \(r \approx 3.004\ \mathrm{cm}\) and \(h \approx 12.377\ \mathrm{cm}\), which are fairly close to the real soda can dimensions of radius \(3.1\ \mathrm{cm}\) and height \(12\ \mathrm{cm}\). Now, considering double-layered top and bottom, modify the surface area formula: \(A'(r) = 3\pi r^2 + 2\pi r \cdot \frac{708}{\pi r^2}\), and repeat steps 3 and 4 to find the new optimal dimensions. Differentiating and solving for the critical points, we get \(r\approx 3.146\ \mathrm{cm}\) and \(h \approx 11.878\ \mathrm{cm}\). The adjusted dimensions of the cylinder are closer to the dimensions of a real soda can, confirming that real cans do take into account factors such as double thickness in their top and bottom surfaces.

Key concepts

Cylindrical Surface Area

When discussing the surface area of a cylinder, it's important to understand that a cylinder consists of a lateral surface and two circular bases. The formula for the total surface area of a cylinder is given by:

\[A = 2\pi r^2 + 2\pi rh\]where:

• \(r\) is the radius of the base of the cylinder
• \(h\) is the height of the cylinder
• The term \(2\pi r^2\) represents the combined area of the top and bottom circles.
• The term \(2\pi rh\) represents the lateral surface area of the cylinder.

To minimize the surface area while maintaining a constant volume, we substitute expressions for volume and leverage calculus techniques such as finding critical points with derivatives. This approach helps in designing shapes like soda cans to use minimal material while containing a given volume.

Volume of a Cylinder

A cylinder's volume gives us an idea of how much space it can hold. The formula to calculate the volume of a cylinder is:

\[V = \pi r^2 h\]In this formula:

• \(r\) is the radius of the cylinder's base
• \(h\) is the height of the cylinder
• \(\pi r^2\) is the area of the base circle

When solving optimization problems involving cylinders, such as reducing material costs or weight, maintaining the volume is crucial. For instance, in the optimization problem, the volume is held constant at 354 cm³, allowing us to express one variable, such as the height, in terms of the other (radius). This dependency is crucial for optimization calculations as it simplifies functions by reducing the number of variables.

Critical Points

Critical points play a pivotal role in optimization problems. They can indicate where maxima and minima of a function occur. To find these points for the function of interest—in our case, the formula for the surface area of a cylinder—we need to differentiate the function and set the derivative to zero.

• In essence, critical points provide the values of a variable (like the radius) where the slope of the function switches direction, which usually marks a peak or a valley.
• For example, the critical point derived in our cylinder problem indicates minimum surface area for a certain configuration.

Next, these points are evaluated to ensure that they minimize or maximize the desired function. In practical scenarios, it helps us design more efficient and less costly objects.

Derivative Applications

Derivatives are powerful tools in calculus that express how a function changes as its input changes. This characteristic makes them invaluable for solving optimization problems.

In the context of a cylindrical shape, derivatives help us find the critical points of the surface area function:

• By finding the derivative of the surface area equation with respect to the radius, we can pinpoint where the surface area is minimized.
• Setting the derivative equal to zero allows us to solve for the radius that provides either a minimum or maximum value of the function, which represents the least use of material in the given problem.
• Finally, by applying second derivative or other test methods, we confirm that the critical points indicate minima (or maxima) in the function.

Understanding and applying derivatives enable efficient design, ensuring that constraints like volume are maintained while optimizing other aspects like surface area.