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Class 12 Applications Of Derivatives · Very Long Answer5 Marksconstrained_optimization

Class 12 Maths · Applications Of Derivatives · Very Long Answer #68

A cuboid has sides x,2x,x/3 and a sphere has radius r, with x,r>0. Their total surface area is fixed at S>0. Show that their total volume is minimized when x=3r and find the minimum in terms of S.

✅ Step-by-Step Solution

Final Answer: x=3r; Vmin=S^(3/2)/[3√(54+4π)].
  1. The total surface area is S=2(2x²+x²/3+2x²/3)+4πr²=6x²+4πr²; total volume is V=2x³/3+4πr³/3.
  2. Write x=√[(S−4πr²)/6], 0<r<√(S/(4π)). Differentiating the constraint gives dx/dr=−2πr/(3x).
  3. Hence dV/dr=2x² dx/dr+4πr²=(4πr/3)(3r−x). It vanishes when x=3r.
  4. As r increases, x decreases, so 3r−x changes once from negative to positive. This proves the global minimum, where r=√[S/(54+4π)].
  5. At x=3r, V=(18+4π/3)r³=(54+4π)r³/3=S^(3/2)/[3√(54+4π)].

📋 CBSE Stepwise Marking Scheme

  • [1 mark] The total surface area is S=2(2x²+x²/3+2x²/3)+4πr²=6x²+4πr²; total volume is V=2x³/3+4πr³/3.
  • [1 mark] Write x=√[(S−4πr²)/6], 0<r<√(S/(4π)). Differentiating the constraint gives dx/dr=−2πr/(3x).
  • [1 mark] Hence dV/dr=2x² dx/dr+4πr²=(4πr/3)(3r−x). It vanishes when x=3r.
  • [1 mark] As r increases, x decreases, so 3r−x changes once from negative to positive. This proves the global minimum, where r=√[S/(54+4π)].
  • [1 mark] At x=3r, V=(18+4π/3)r³=(54+4π)r³/3=S^(3/2)/[3√(54+4π)].

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