Question:hard

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

Show Hint

Write the volume as a function of one variable using the given sum, then apply calculus (or AM-GM) to maximize it.
Updated On: Jul 21, 2026
  • \(\dfrac{256\pi}{9}\) cc
  • \(\dfrac{512\pi}{9}\) cc
  • \(\dfrac{256\pi}{27}\) cc
  • \(\dfrac{512\pi}{27}\) cc
Show Solution

The Correct Option is D

Solution and Explanation

This can also be solved without calculus, using the AM-GM inequality.
Step 1: Set up the constraint. With diameter plus height equal to 8 cm, \(2r + h = 8\), which can be written as \(r + r + h = 8\).
Step 2: Apply AM-GM to the three terms r, r, h. By AM-GM, \(\dfrac{r+r+h}{3} \geq \sqrt[3]{r \times r \times h}\), so \(\dfrac{8}{3} \geq \sqrt[3]{r^2 h}\).
Step 3: Cube both sides. \(\left(\dfrac{8}{3}\right)^3 \geq r^2h\), that is \(r^2h \leq \dfrac{512}{27}\).
Step 4: Find when equality holds. AM-GM gives equality when all three terms are equal, that is when \(r = h\). Substituting into \(2r+h=8\) with \(r=h\) gives \(3r = 8\), so \(r = h = \dfrac{8}{3}\).
Step 5: State the maximum volume. Since \(r^2h\) reaches its maximum value of \(512/27\) at this point, the maximum volume is \(V = \pi r^2 h = \dfrac{512\pi}{27}\) cc, matching the calculus result.\[\boxed{\dfrac{512\pi}{27}\text{ cc}}\]
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