Problem 37 - Calculus Early Transcendentals

Question

A cone-shaped paper drinking cup is to hold \(100 \mathrm{cm}^{3}\) of water. Find the height and radius of the cup that will require the least amount of paper.

Step-by-Step Solution

Answer

Find the critical points of the function for surface area in terms of the radius, solve for the radius, and use it to find the height.

Step 1: Express Volume Constraint

The volume of a cone is given by the formula \( V = \frac{1}{3} \pi r^2 h \). Given that the volume is \(100 \, \text{cm}^3\), we have the equation \( \frac{1}{3} \pi r^2 h = 100 \). This is our volume constraint.

Step 2: Express Surface Area to Minimize

The surface area of the cone (the amount of paper needed) is \( A = \pi r l \), where \( l = \sqrt{r^2 + h^2} \) is the slant height. Our goal is to minimize this surface area.

Step 3: Substitute Height from Volume Constraint

From the volume constraint \( \frac{1}{3} \pi r^2 h = 100 \), solve for \( h \): \( h = \frac{300}{\pi r^2} \). Substitute this expression for \( h \) into the surface area equation to express \( A \) only in terms of \( r \).

Step 4: Express Surface Area Function

Substituting \( h \) in the expression for slant height gives \( l = \sqrt{r^2 + \left(\frac{300}{\pi r^2}\right)^2} \). Now, substitute \( l \) in \( A = \pi r l \) to get a function \( A(r) \) that we can minimize.

Step 5: Differentiate and Find Critical Points

Find the derivative \( \frac{dA}{dr} \) and set it to zero to find critical points. Calculating the derivative of the complex expression found in Step 4 might involve using the chain rule and square root derivatives.

Step 6: Solve for Optimal Radius

Solve the equation \( \frac{dA}{dr} = 0 \) for \( r \) to find the critical points. Confirm which value of \( r \) is a minimum by evaluating the second derivative \( \frac{d^2A}{dr^2} \), or by testing values.

Step 7: Find Corresponding Height

Use the optimal \( r \) found in Step 6 in the volume constraint \( h = \frac{300}{\pi r^2} \) to find the corresponding height \( h \).

Step 8: Verify Solution

Check if the values satisfy both the volume constraint and minimize the surface area calculated in earlier steps.

Key Concepts

Cone Volume Calculation

When working with problems involving cones, understanding the calculation of their volume is key. The volume formula for a cone is given by:\[V = \frac{1}{3} \pi r^2 h\]This formula represents the volume, where:

• \( V \) is the volume,
• \( r \) is the radius of the base,
• \( h \) is the height of the cone.

In this particular problem, the volume is fixed at \(100 \, \mathrm{cm}^3\). By plugging this value into the equation, students can set up an equation relating the radius and the height of the cone. This volume equation will be pivotal when transforming the surface area formula to allow calculations solely in terms of one variable.

Surface Area Minimization

The objective of our problem is to minimize the surface area of a cone-shaped paper cup, which practically means using the least amount of paper. The formula for the lateral surface area of a cone is:\[A = \pi r l\]Here,

• \( A \) is the surface area,
• \( r \) is the radius of the base,
• \( l \) is the slant height of the cone.

The slant height \( l \) is calculated using the Pythagorean theorem, giving us \( l = \sqrt{r^2 + h^2} \). This creates a relationship between the radius, height, and slant height. By substituting the expression of \( h \) from the volume equation, we can express \( l \) and hence the surface area \( A \) purely in terms of the radius \( r \). This transformation is crucial for applying calculus-based optimization techniques.

Critical Points in Calculus

To find the minimum surface area in this problem, students need to determine the critical points of the surface area function. Critical points occur where the derivative of a function is zero or undefined. For our problem:- Begin by differentiating the surface area function, \( A(r) \), with respect to \( r \).- This involves using the chain rule and product rule, owing to the square root in the slant height expression.- Set the derivative \( \frac{dA}{dr} \) equal to zero to locate the critical points.Finding these points helps identify where the minimum surface area occurs. After finding the critical points, it is essential to determine if they correspond to a minimum by evaluating the second derivative or using other testing methods.

Optimization Techniques

Optimization in calculus involves finding the minimum or maximum values of functions within constraints. This exercise uses optimization to minimize the surface area of a cone while keeping the volume constant. Here’s a brief run-through of the typical optimization process demonstrated in this exercise:

• **Define Constraints**: Start with the volume constraint equation.
• **Express Objective Function**: Create an expression for the surface area, integrating the constraints.
• **Differentiate and Find Critical Points**: Find the derivative of the surface area function and solve for critical points.
• **Analyze Critical Points**: Use the second derivative test or test values to identify and confirm the minimum.
• **Verify Solutions**: Check that solutions meet the initial conditions and constraints.

These steps illustrate the techniques and considerations in optimization problems, which are fundamental in calculus to solve practical real-world issues.