1.

Prove that the height of a cylinder of maximum volume, inscribed insphere of radius 'a' is 2a/V3.

Answer»

Let r be the radius and h the height of the inscribed cylinder ABCD.let V be its volume.then , V = πr²h------( 2 )clearly, AC = 2R.ALSO, AC² = AB² + BC²

=> (2R)² = (2r)² + h²=> r² = 1/4(4R² - h²)--------( 2 )

using----( 1 ) & -----( 2 )V = πh/4(4R² - h²)=> dV/dh = (πR² - 3/4πh²)

and d²V/dh² = -3/2πh.

for Maxima or minima, we have (dV/dh) = 0

now, dV/dh = 0=> πR² - 3/4πh² = 0h = 2R/√3.

therefore,[d²V/dh²](at h = 2R/√3) = -3/2π * 2R/√3= -πR√3 < 0.

So, V is maximum when h = 2R/√3.hence, the height of the cylinder of maximum volume is 2R/√3.Largest volume of the cylinder = π×1/4[4R² - 4R²/3] × 2R/√3 = 4πR³/3√3.


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