Question

If the volume of a right circular cone is given, then its curved surface area is minimum when the ratio of its height to its base radius,

Answer 1

Let rr be the radius & hh be the height of the cylinder having its total surface area AA(constant) since cylindrical container is closed at the top (circular) then its surface area (constant\fixed) is given as

=(area of lateral surface)+2(area of circular top/bottom)=(area of lateral surface)+2(area of circular top/bottom)

A=2πrh+2πr2A=2πrh+2πr2

h=A−2πr22πr=A2πr−r(1)(1)h=A−2πr22πr=A2πr−r

Now, the volume of the cylinder

V=πr2h=πr2(A2πr−r)=A2r−πr3V=πr2h=πr2(A2πr−r)=A2r−πr3

differentiating VV w.r.t. rr, we get

dVdr=A2−3πr2dVdr=A2−3πr2

d2Vdr2=−6πr<0  (∀  r>0)d2Vdr2=−6πr<0  (∀  r>0)

Hence, the volume is maximum, now, setting dVdr=0dVdr=0 for maxima

A2−3πr2=0⟹r=A6π−−−√A2−3πr2=0⟹r=A6π

Setting value of rr in (1), we get

h=A2πA6π−−√−A6π−−−√=(32−−√−16–√)Aπ−−√=2A3π−−−√h=A2πA6π−A6π=(32−16)Aπ=2A3π

Hence, the ratio of height (h)(h) to the radius (r)(r) is given as

h/r =2
hr=2A3π−−−√A6π−−√=12πA3πA−−−−−√=2hr=2A3πA6π=12πA3πA=2