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Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #101

Show that an open-top right circular cylinder of fixed positive volume has least total material surface area when its height equals its radius.
Open cylinder: axial section
Open cylinder: axial section

✅ Step-by-Step Solution

Final Answer: h=r=(V/π)^(1/3).
  1. For fixed V>0, the constraint V=πr²h gives h=V/(πr²), r>0.
  2. The base plus curved area is S=πr²+2πrh=πr²+2V/r.
  3. S′=2πr−2V/r²=0 implies πr³=V.
  4. S″=2π+4V/r³>0, so the critical radius is the unique global minimum.
  5. Then h=V/(πr²)=r=(V/π)^(1/3), as required.

📋 CBSE Stepwise Marking Scheme

  • [1 mark] For fixed V>0, the constraint V=πr²h gives h=V/(πr²), r>0.
  • [1 mark] The base plus curved area is S=πr²+2πrh=πr²+2V/r.
  • [1 mark] S′=2πr−2V/r²=0 implies πr³=V.
  • [1 mark] S″=2π+4V/r³>0, so the critical radius is the unique global minimum.
  • [1 mark] Then h=V/(πr²)=r=(V/π)^(1/3), as required.

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