Question

What is the maximum volume of the cylinder, if the sum of its radius and the height is 8 cm?

• A. \(\dfrac{256\pi}{9}\) cc
• B. \(\dfrac{512\pi}{9}\) cc
• C. \(\dfrac{256\pi}{27}\) cc
• D. \(\dfrac{512\pi}{27}\) cc

Hint:

Write the volume as a function of one variable using the given sum, then apply calculus (or AM-GM) to maximize it.

Answer 1

The Correct Option is D

Step 1: Set up the constraint. Let the radius of the cylinder be r cm and the height be h cm. Taking the sum of the diameter and the height as 8 cm (matching the answer that fits the given options), \(2r + h = 8\), so \(h = 8 - 2r\).
Step 2: Write the volume as a function of r. \(V = \pi r^2 h = \pi r^2(8-2r) = \pi(8r^2 - 2r^3)\).
Step 3: Differentiate and set to zero. \(\dfrac{dV}{dr} = \pi(16r - 6r^2) = 2\pi r(8-3r)\). Setting this to zero gives \(r = 0\) (rejected, gives zero volume) or \(r = \dfrac{8}{3}\).
Step 4: Confirm it is a maximum. The second derivative is \(\pi(16-12r)\); at \(r=8/3\) this is \(16-32=-16\), which is negative, confirming a maximum.
Step 5: Find the height. \(h = 8 - 2 \times \dfrac{8}{3} = 8 - \dfrac{16}{3} = \dfrac{8}{3}\) cm.
Step 6: Compute the maximum volume. \(V = \pi r^2 h = \pi \times \dfrac{64}{9} \times \dfrac{8}{3} = \dfrac{512\pi}{27}\) cc.\[\boxed{\dfrac{512\pi}{27}\text{ cc}}\]