Question

Refer the attached picture. ​

Answer 1

Let S be the given surface area of the closed cylinder whose radius is r and height h.

Let V be its volume.

Surface Area S=2πr2+2πrh

h=S−2πr22πr----(1)

Volume V=πr2h

Substitute the value of h in the above equation

=πr2[S−2πr22πr]

=12r(S−2πr)

=12[Sr−2πr3]

Differentiating with respect to r we get

dVdx=12[S−6πr2]

Step 2:

For maxima and minima dVdx=0

S−6πr2=0

S=6πr2

From (1)

h=S−2πr22πr

Substitute the value os S in the above equation

h=6πr2−2πr22πr

=4πr22πr

=2r

h=2r

Step 3:

On double differentiation of V we get,

d2Vdr2=12(−12πr)

=−6πr

=−ve

∴V is maximum

Thus volume is maximum when h=2r

(i.e)when height of cylinder =diameter of the base.

Hope helps