Let r and h be the radius and height of cone inscribed in sphere of radius 12cm
In △AOB,122=(h−12)2+r2
∴ r2=24h−h2
volume,
v=31πr2h
v=31π(24h−h2)h
∴v=3π(24h2−h3)
derivating the above equation, we get,
dhdv=3π(48h−3h2)
equate dhdv=0
⇒ 3π(48h−3h2)=0
solving the above equation, we get,
h=16
Again derivating the above equation, we get,
dh2d2v=3π(48−6h)
dh2d2v→h=16=3π(48−6(16))<0
Volume is maximum, when h=16
Hence the height of the cone