Question:medium

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

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Write the volume as a single-variable function using the given sum, differentiate and set to zero to find the optimum radius, then substitute back to get the maximum volume.
Updated On: Jul 20, 2026
  • \( \frac{256\pi}{9} \) cc
  • \( \frac{512\pi}{9} \) cc
  • \( \frac{256\pi}{27} \) cc
  • \( \frac{512\pi}{27} \) cc
  • \( \frac{1024\pi}{27} \) cc
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The Correct Option is D

Solution and Explanation

This same optimisation can be done without calculus, using the AM-GM inequality, working with diameter + height = 8 (the reading that reproduces one of the given options).

Let radius $=r$ and height $=h$, with $2r+h=8$.
We want to maximise $V=\pi r^2h$.

Write $2r+h$ as a sum of three equal-weighted terms: $r+r+h=8$.
By AM-GM on $r, r, h$:
$$\frac{r+r+h}{3}\ge\sqrt[3]{r\cdot r\cdot h}$$
$$\frac{8}{3}\ge\sqrt[3]{r^2h}$$
Cubing both sides:
$$\left(\frac{8}{3}\right)^3\ge r^2h$$
$$\frac{512}{27}\ge r^2h$$
Equality (the maximum) holds when all three terms are equal, i.e. $r=h$.

Substituting $r=h$ into $2r+h=8$: $2r+r=8 \Rightarrow r=\frac{8}{3}$, so $h=\frac{8}{3}$ too.

Maximum of $r^2h = \frac{512}{27}$, so:
$$V_{max}=\pi\times\frac{512}{27}=\frac{512\pi}{27}\text{ cc}$$

This confirms the same value obtained via calculus under the diameter + height = 8 reading.
\[\boxed{\frac{512\pi}{27}\text{ cc}}\]
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