Question

Surface area of closed
100 sq.om ST its vol is maximum when
base radius is half the height of
cylinder by differentiation method​

Answer 1

Step-by-step explanation:

Let r be the radius & h be the height of the cylinder having its total surface area A (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)

A=2πrh+2πr2

h=A−2πr22πr=A2πr−r(1)

Now, the volume of the cylinder

V=πr2h=πr2(A2πr−r)=A2r−πr3

differentiating V w.r.t. r, we get

dVdr=A2−3πr2

d2Vdr2=−6πr<0 (∀ r>0)

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

A2−3πr2=0⟹r=A6π−−−√

Setting value of r in (1), we get

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

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

hr=2A3π−−−√A6π−−√=12πA3πA−−−−−√=2

hr=2