Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Of all right circular cylinders with a given surface area, find the one with the maximum volume. Note: The ends of the cylinders are closed.

Step-by-Step Solution

Answer

The cylinder with maximum volume has radius and height both equal to \( \sqrt{\frac{S}{6\pi}} \).

Step 1: Understand the Problem

We need to find the dimensions of a right circular cylinder that maximize its volume while keeping its surface area constant. The ends of the cylinder are closed.

Step 2: Define Variables and Formulas

Let the radius of the base of the cylinder be \( r \) and the height of the cylinder be \( h \). The surface area \( S \) of the cylinder is given by the formula \( S = 2\pi r^2 + 2\pi rh \). The volume \( V \) of the cylinder is given by \( V = \pi r^2 h \).

Step 3: Express One Variable in Terms of Another

Using the surface area formula, express \( h \) in terms of \( r \): \( S = 2\pi r^2 + 2\pi rh \Rightarrow h = \frac{S - 2\pi r^2}{2\pi r} \).

Step 4: Substitute into the Volume Formula

Substitute the expression for \( h \) into the volume formula: \( V = \pi r^2 \left( \frac{S - 2\pi r^2}{2\pi r} \right) = \frac{r(S - 2\pi r^2)}{2} \).

Step 5: Differentiate to Find Critical Points

Differentiate the volume function with respect to \( r \): \( \frac{dV}{dr} = \frac{S - 6\pi r^2}{2} \). Set \( \frac{dV}{dr} = 0 \) to find critical points: \( S - 6\pi r^2 = 0 \Rightarrow r = \sqrt{\frac{S}{6\pi}} \).

Step 6: Verify the Maximum Volume

Using the second derivative test, calculate \( \frac{d^2V}{dr^2} = -6\pi r \). Since \( \frac{d^2V}{dr^2} < 0 \) for \( r = \sqrt{\frac{S}{6\pi}} \), this critical point is a maximum.

Step 7: Calculate Corresponding Height

Substitute \( r = \sqrt{\frac{S}{6\pi}} \) back into the expression for \( h \): \( h = \frac{S - 2\pi \left(\sqrt{\frac{S}{6\pi}}\right)^2}{2\pi \sqrt{\frac{S}{6\pi}}} = \sqrt{\frac{S}{6\pi}} \).

Step 8: Conclusion

The right circular cylinder with maximum volume for a given surface area has equal height and radius, both equal to \( \sqrt{\frac{S}{6\pi}} \).

Key Concepts

Volume of a Cylinder

When studying the geometry of a right circular cylinder, the volume calculation is crucial. The cylinder is defined by two parameters: its radius \( r \) and its height \( h \). The volume \( V \) of the cylinder can be expressed by the formula:

• \( V = \pi r^2 h \)

This formula is derived from the base area of the circular face \( \pi r^2 \) multiplied by the height \( h \). Maximizing the volume of a cylinder is often an optimization challenge, particularly when constraints like surface area are involved. The volume depends directly on both the radius and height, which is why it's essential to consider any given constraints when calculating to ensure the values are feasible and optimal for the given problem.

Surface Area Constraint

For this problem, the surface area of the cylinder is the constraint we must work under. The surface area \( S \) includes the areas of the two circular ends and the lateral surface, and it is given by the formula:

• \( S = 2\pi r^2 + 2\pi rh \)

This formula combines:

• the area of both circular ends: \( 2\pi r^2 \)
• the lateral area: \( 2\pi rh \)

The challenge is to express one of the dimensions, usually the height \( h \), in terms of the other, the radius \( r \), using this surface area constraint. By doing so, we can rewrite the volume formula solely in terms of \( r \), simplifying the optimization process. In this problem, solving for \( h \) involved rearranging the surface area formula to find:

• \( h = \frac{S - 2\pi r^2}{2\pi r} \)

Using such a constraint-driven approach ensures that while maximizing the volume, we respect the limits set by the surface area.

Critical Points and Second Derivative Test

Optimization in calculus often involves finding the critical points of a function. To do this, we differentiate the function with respect to the variable. In this scenario, after expressing the volume \( V \) solely in terms of \( r \), the derivative \( \frac{dV}{dr} \) is obtained:

• \( \frac{dV}{dr} = \frac{S - 6\pi r^2}{2} \)

Setting this derivative equal to zero finds the critical points:

• \( S - 6\pi r^2 = 0 \)
• So, \( r = \sqrt{\frac{S}{6\pi}} \)

Once a critical point is identified, it's essential to determine whether it represents a maximum, minimum, or saddle point. This is done using the second derivative test. Calculating the second derivative of the volume function gives:

• \( \frac{d^2V}{dr^2} = -6\pi r \)

With \( \frac{d^2V}{dr^2} < 0 \) for \( r = \sqrt{\frac{S}{6\pi}} \), it confirms the critical point is a maximum. Therefore, the dimensions \( r \) and \( h \) that maximize the cylinder's volume are equal, both at \( \sqrt{\frac{S}{6\pi}} \), showcasing the power of calculus in solving real-world optimization problems.