A rectangle of perimeter \(24\text{ cm}\) is revolved along one of its sides to sweep out a cylinder of maximum volume. Find the dimensions of the rectangle.
Show Hint
Pay close attention to which side the rectangle is revolved around! Revolving around side \(y\) makes the volume proportional to \(x^2 y\), whereas revolving around side \(x\) would make the volume proportional to \(y^2 x\). This choice changes which variable is squared in the volume equation.
Concept:
When a rectangle with adjacent side lengths \(x\) and \(y\) is revolved around one of its sides, say the side of length \(y\), that side becomes the fixed vertical height (\(h\)) of the resulting cylinder. The other perpendicular side of length \(x\) revolves to sweep out a circular base, forming the radius (\(r\)) of the cylinder.
• Volume of a cylinder: \(V = \pi r^2 h\)
• Perimeter of a rectangle: \(P = 2(x + y)\)
To maximize the volume function, we express it in terms of a single variable using the perimeter relationship, find its first derivative, set it to zero to locate critical points, and apply the second derivative test to confirm it is a local maximum.
Step 1: Set up the equations for perimeter and cylinder volume.
Let the side acting as the radius of the cylinder be \(x\), and the side acting as the height of the cylinder be \(y\).
The perimeter of the rectangle is given as \(24\text{ cm}\):
\[
2(x + y) = 24 \quad \Rightarrow \quad x + y = 12 \quad \Rightarrow \quad y = 12 - x \quad \cdots (1)
\]
When the rectangle is revolved around the side of length \(y\), the dimensions of the generated cylinder are:
\[
\text{Radius } (r) = x \quad \text{and} \quad \text{Height } (h) = y = 12 - x
\]
The formula for the volume \(V\) of a cylinder is:
\[
V = \pi r^2 h
\]
Substituting the expressions for \(r\) and \(h\) in terms of \(x\):
\[
V(x) = \pi x^2 (12 - x)
\]
Expanding the expression:
\[
V(x) = \pi (12x^2 - x^3) \quad \cdots (2)
\]
Step 2: Find the critical points using the first derivative.
To find the dimensions that maximize the volume, differentiate \(V(x)\) with respect to \(x\):
\[
\frac{dV}{dx} = \pi \frac{d}{dx}(12x^2 - x^3)
\]
\[
\frac{dV}{dx} = \pi (24x - 3x^2)
\]
Set the first derivative equal to zero to find the critical points:
\[
\pi (24x - 3x^2) = 0
\]
Since \(\pi \neq 0\), we can factor the remaining terms:
\[
3x(8 - x) = 0
\]
This gives two mathematical solutions:
\[
x = 0 \quad \text{or} \quad x = 8
\]
Since a geometric side length must be greater than zero, \(x = 0\) is discarded. Thus, our valid critical point is \(x = 8\).
Step 3: Verify the maximum using the second derivative test.
Let us find the second derivative of the volume function with respect to \(x\):
\[
\frac{d^2V}{dx^2} = \pi \frac{d}{dx}(24x - 3x^2) = \pi (24 - 6x)
\]
Now, evaluate this second derivative at our critical point \(x = 8\):
\[
\left. \frac{d^2V}{dx^2} \right|_{x=8} = \pi (24 - 6(8)) = \pi (24 - 48) = -24\pi
\]
Since the second derivative is negative (\(-24\pi < 0\)), the volume reaches a local maximum at \(x = 8\).
Step 4: Compute the final dimensions of the rectangle.
Substitute the value \(x = 8\text{ cm}\) back into equation (1) to find the remaining side dimension \(y\):
\[
y = 12 - 8 = 4\text{ cm}
\]
Thus, the dimensions of the rectangle required to maximize the swept volume are a length of \(8\text{ cm}\) and a width of \(4\text{ cm}\). This matches option (B).
