Problem 30 - Calculus Early Transcendentals

Question

A closed cylindrical can is to have a surface area of \(S\) square units. Show that the can of maximum volume is achieved when the height is equal to the diameter of the base.

Step-by-Step Solution

Answer

The height equals the diameter of the base for maximum volume.

Step 1: Understand the Problem

We need to find the relationship between the height and diameter that maximizes the volume of a closed cylindrical can, given a fixed surface area, denoted as \( S \).

Step 2: Write the Formulas

The surface area \( S \) of a closed cylinder with radius \( r \) and height \( h \) is given by: \[ S = 2\pi r^2 + 2\pi rh \] The volume \( V \) of the cylinder is given by: \[ V = \pi r^2 h \]

Step 3: Express Height in Terms of the Surface Area and Radius

From the equation for surface area, solve for \( h \): \[ S = 2\pi r^2 + 2\pi rh \] \[ h = \frac{S - 2\pi r^2}{2\pi r} \]

Step 4: Substitute the Height Expression into the Volume Formula

Substitute the expression for \( h \) from Step 3 into the volume formula:\[ V = \pi r^2 \left(\frac{S - 2\pi r^2}{2\pi r}\right) = \frac{r(S - 2\pi r^2)}{2} \]Simplifying gives:\[ V = \frac{Sr - 2\pi r^3}{2} \]

Step 5: Find the Derivative of the Volume

Differentiate \( V \) with respect to \( r \):\[ \frac{dV}{dr} = \frac{S}{2} - 3\pi r^2 \]

Step 6: Set the Derivative Equal to Zero to Find Critical Points

Set \( \frac{dV}{dr} = 0 \):\[ \frac{S}{2} - 3\pi r^2 = 0 \]Solving for \( r \) yields:\[ 3\pi r^2 = \frac{S}{2} \]\[ r = \sqrt{\frac{S}{6\pi}} \]

Step 7: Determine the Relationship Between Height and Diameter

Now calculate the height \( h \) when \( r = \sqrt{\frac{S}{6\pi}}\) using the expression from Step 3:\[ h = \frac{S - 2\pi \left(\sqrt{\frac{S}{6\pi}}\right)^2}{2\pi \sqrt{\frac{S}{6\pi}}} \]It turns out:\[ h = 2r \] or \( h = 2 \left(\frac{S}{6\pi}\right)^{1/2}\), which implies \( h\) equals the diameter.

Key Concepts

Volume Maximization

When optimizing the volume of a cylinder with a fixed surface area, our goal is to find the dimensions that give us the maximum volume. A closed cylinder has two components contributing to volume:

• The base, defined by its radius \( r \)
• The height \( h \)

The volume \( V \) of a cylinder is given by the formula:\[V = \pi r^2 h\]This equation shows that increasing either the base area \( \pi r^2 \) or the height \( h \) increases the overall volume of the cylinder. However, with a surface area constraint, we must balance the relationship between \( r \) and \( h \) to maximize \( V \). This delicate balance involves adjusting one dimension without compromising the surface area, leading to an optimal size configuration for maximum volume.

Surface Area Constraint

A surface area constraint is a boundary condition which restricts how the dimensions of the cylinder can be changed while optimizing its volume. For a closed cylinder, the surface area \( S \) is defined by:\[S = 2\pi r^2 + 2\pi rh\]This formula represents:

• The area of the top and bottom circles: \( 2\pi r^2 \)
• The area of the side of the cylinder: \( 2\pi rh \)

Since the total surface area \( S \) is fixed, changes to either the radius \( r \) or height \( h \) must maintain the equation's balance. Therefore, maximizing volume with fixed surface area means reconfiguring \( r \) and \( h \) such that the equation still holds true. Solving for one dimension as a function of the other typically helps in eliminating variables and simplifying the optimization process.

Derivative Test

Using calculus, we apply the derivative test to identify points where the volume function reaches a maximum or minimum. We begin by expressing the volume \( V \) in terms of a single variable, often the radius \( r \), after substituting the expression for \( h \) from the surface area constraint. This creates a function of one variable:\[V = \frac{Sr - 2\pi r^3}{2}\]To find these extrema, we compute the derivative \( \frac{dV}{dr} \):\[\frac{dV}{dr} = \frac{S}{2} - 3\pi r^2\]Setting the derivative to zero helps us locate critical points where the function may reach a maximum or minimum. Solving\[\frac{S}{2} - 3\pi r^2 = 0\]gives us critical points that are candidates for the volume's maximal configuration.

Critical Points

Critical points are the values of the variable, in this case, the radius \( r \), where the derivative becomes zero or does not exist. These points indicate potential maxima or minima of the volume. After finding the critical point:\[r = \sqrt{\frac{S}{6\pi}}\]we then substitute back to find the corresponding height \( h \). This involves utilizing the earlier expression for \( h \):\[h = \frac{S - 2\pi \left(\sqrt{\frac{S}{6\pi}}\right)^2}{2\pi \sqrt{\frac{S}{6\pi}}}\]Solving gives:\[h = 2r\]This result leads to the essential insight of the problem: the height of the cylinder is twice the radius, meaning when both height and diameter are equal, the cylinder's volume is optimized within the surface area constraints.