Question

Show that the cylinder of given volume open at the top has minimum total surface area if its height is equal to radius of the base.

Answer 1

Let rr be the radius and hh be the the height of a cylinder of given volume VV

V=πr2hV=πr2h

h=vπr2h=vπr2

Let SS be total surface area.Since the cylinder is open at its top.

S=2πrh+πr2S=2πrh+πr2

Substituting for hh we get,

S=2πr(vπr2)S=2πr(vπr2)+πr2+πr2

=2vr=2vr+πr2+πr2

Differentiating with respect to xx we get,

dSdr=−2vr2dSdr=−2vr2+2πr+2πr

For maximum or minimum

dSdr=dSdr=00

⇒2vr2+⇒2vr2+2πr=02πr=0

(i.e) −vr2−vr2+πr=0+πr=0

∴v=πr3∴v=πr3

πr2h=πr3πr2h=πr3

⇒h=r⇒h=r

Differentiating again with respect to rr we get,

d2Sdr2=4vr3d2Sdr2=4vr3+2π+2π

When r=hr=h

d2Sdr2=4vr3d2Sdr2=4vr3+2π+2π >0

Hence SS is minimum when h=rh=r (i.e) when the height of the cylinder is equal to the radius of the base.