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NCERT Class 12 · Ch 6 Miscellaneous Exercise · Q15

NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Miscellaneous Exercise Question 15

A cylinder is inscribed coaxially in a right circular cone of height H and semi-vertical angle α. Show its maximum-volume height is H/3 and Vmax=(4/27)πH³tan²α.
Optimal cylinder in cone: representative H=3, tan α=1
Optimal cylinder in cone: representative H=3, tan α=1

💡 Concept Hints

  1. Identify the domain and the inner and outer functions, or the relevant one-sided limits.
  2. Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.

✅ Step-by-Step Solution

Final Answer: h=H/3; Vmax=4πH³tan²α/27.
  1. For cylinder height h from the cone base, similar triangles give cylinder radius r=(H−h)tan α, 0<h<H.
  2. Thus V=πtan²α·h(H−h)². Its derivative is πtan²α(H−h)(H−3h).
  3. The only feasible interior stationary point is h=H/3, with derivative changing + to − and endpoint volumes zero.
  4. At that height r=(2H/3)tan α and Vmax=4πH³tan²α/27.

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