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NCERT Class 12 · Ch 6 Exercise 6.3 · Q20

NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 20

Show that a right circular cylinder with fixed surface area and maximum volume has height equal to its base diameter.
Optimal cylinder: axial section h=2r
Optimal cylinder: axial section h=2r

💡 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=2r (closed cylinder, total surface area fixed).
  1. For fixed total surface S=2πr²+2πrh, h=S/(2πr)−r. Positive height requires 0<r<√(S/(2π)).
  2. V=πr²h=Sr/2−πr³, so V′=S/2−3πr². The stationary radius has S=6πr².
  3. Then h=S/(2πr)−r=2r. V″=−6πr<0, and volumes approach zero at the domain endpoints.

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