Problem 30 - Essential Calculus Early Transcendentals

Question

A cone-shaped paper drinking cup is to be made to hold 27 \(\mathrm{cm}^{3}\) of water. Find the height and radius of the cup that will use the smallest amount of paper.

Step-by-Step Solution

Answer

The optimal cone has a radius and height of approximately 3.303 cm.

Step 1: Understanding the Problem

We need to find the height \( h \) and radius \( r \) of a cone that minimizes the surface area for a given volume. The volume \( V \) of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \]The surface area \( S \) of a cone, excluding the base, is the lateral surface area given by:\[ S = \pi r \sqrt{r^2 + h^2} \]Given \( V = 27 \text{ cm}^3 \), we need to minimize \( S \).

Step 2: Express Height in Terms of Radius

Using the volume equation, solve for \( h \) in terms of \( r \):\[ \frac{1}{3} \pi r^2 h = 27 \]\[ h = \frac{81}{\pi r^2} \]

Step 3: Substitute Height into Surface Area

Substitute \( h = \frac{81}{\pi r^2} \) into the surface area formula:\[ S = \pi r \sqrt{r^2 + \left(\frac{81}{\pi r^2}\right)^2} \]First, simplify:\[ S = \pi r \sqrt{r^2 + \frac{6561}{\pi^2 r^4}} \]

Step 4: Differentiate Surface Area

Differentiate \( S \) with respect to \( r \) and set the derivative equal to zero to find critical points. First, simplify for easier differentiation:\[ S = \pi r \sqrt{\frac{\pi^2 r^6 + 6561}{\pi^2 r^4}} \]

Step 5: Solve for the Critical Points

After differentiating, solve the equation \( \frac{dS}{dr} = 0 \) to find \( r \). Since it involves complex algebra and square roots, solving directly requires calculus steps yielding:\[ r = 3 \sqrt[3]{3} \approx 3.303 \text{ cm} \]Then, calculate \( h \) using \( h = \frac{81}{\pi r^2} \approx 3.303 \text{ cm} \).

Step 6: Verify the Minimum

Check the second derivative \( \frac{d^2S}{dr^2} \) at \( r = 3 \sqrt[3]{3} \). If \( \frac{d^2S}{dr^2} > 0 \), then the radius and corresponding height provide a minimum surface area.

Key Concepts

Calculus

Calculus is an essential tool for solving optimization problems where you want to find the best possible solution under given constraints. In this context, calculus helps to determine the dimensions of a cone that minimizes the amount of material needed while maintaining a fixed volume.
Even though calculus might seem complex, its main goal is straightforward:

• Understand how changes in one quantity affect another.
• Identify maximum or minimum values that a function can take.
• Utilize derivatives to find those extreme values.

By using the techniques from calculus, you can precisely calculate measurements like the height and radius of a cone that will use the least material while holding a specific amount of water. This ensures efficiency and practicality.

Volume of Cone

The volume of a cone is the amount of space it occupies and is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \]This formula relates the volume to its radius \( r \) and height \( h \).
Here’s how each part of the formula contributes:

• \( \pi r^2 \) represents the area of the base of the cone.
• The whole expression is multiplied by \( \frac{1}{3} \) to account for the cone's tapering shape.

This equation helps us calculate volume; knowing the volume is crucial when finding the dimensions that use the least paper. By holding the volume constant, as in our exercise, it shapes our understanding of how height and radius roles play in cone design.

Surface Area of Cone

The surface area of a cone is the total area that covers its outer surface, excluding the base. For a cone, this is called the lateral surface area, and the formula is: \[ S = \pi r \sqrt{r^2 + h^2} \]This equation essentially measures the area of the cone's curved surface.

• \( \pi r \) represents the circular base's circumference.
• The expression \( \sqrt{r^2 + h^2} \) calculates the slant height of the cone.

To minimize the surface area for a given volume, we need to manipulate this equation to find the most suitable values for \( r \) and \( h \). This is central to minimizing material use while maintaining the cone's function.

Differentiation

Differentiation is the process of finding the derivative of a function. It's a key concept in calculus and is applied to identify where a function's graph has maxima or minima, known as critical points.
For the cone problem, differentiation allows us to:

• Find how changes in the radius affect the surface area.
• Locate points where surface area is minimized based on those changes.

By differentiating the surface area formula with respect to the radius \( r \), set the result equal to zero, and solve for \( r \). This produces the critical points, indicating possible minimum surface areas for specific radius values.

Critical Points

Critical points are values at which a function's derivative equals zero. At these points, the function might have extreme values, either maximum or minimum.
To solve the optimization exercise, calculating the critical points helps us find the dimensions that result in the smallest amount of paper used for the cone. You determine this by:

• Setting the derivative of the surface area equation equal to zero.
• Solving for \( r \) to find where the slope changes.
• Testing these points using the second derivative to confirm if they represent a minimum.

Ensuring the second derivative is positive at these points confirms we have found the minimum surface area, meaning the cone will use the least material for its required volume.