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Class 12 Applications Of Derivatives · Very Long Answer5 Marksgeometric_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #70

Of all closed right circular cylindrical cans of volume 128π cm³, find the dimensions with minimum total surface area.
Closed cylinder: axial section
Closed cylinder: axial section

✅ Step-by-Step Solution

Final Answer: Radius 4 cm, height 8 cm; minimum area 96π cm².
  1. For radius r>0, volume πr²h=128π gives h=128/r².
  2. S=2πrh+2πr²=256π/r+2πr².
  3. S′=−256π/r²+4πr=0 gives r³=64, hence r=4.
  4. S″=512π/r³+4π>0, so S is strictly convex and the stationary point is its global minimum.
  5. h=128/16=8 cm, with Smin=256π/4+2π·16=96π cm².

📋 CBSE Stepwise Marking Scheme

  • [1 mark] For radius r>0, volume πr²h=128π gives h=128/r².
  • [1 mark] S=2πrh+2πr²=256π/r+2πr².
  • [1 mark] S′=−256π/r²+4πr=0 gives r³=64, hence r=4.
  • [1 mark] S″=512π/r³+4π>0, so S is strictly convex and the stationary point is its global minimum.
  • [1 mark] h=128/16=8 cm, with Smin=256π/4+2π·16=96π cm².

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