A solid is formed by attaching a hemisphere to each end of a cylinder. If the total volume is to be 120cm^3, find the radius (in cm) of the cylinder that produces the minimum surface area. Express your answer correct to 2 decimal places. Help!?

1 Answer
Aug 14, 2017

#r = ((3V)/(4pi))^(1/3)#.

Explanation:

Assume without loss of generality the cylinder has length #l#, the hemispheres and cylinder have radius #r#, and the entire geometric object has volume #V# and surface area S.

Two hemispheres attached to either end have the equivalent volume of a single sphere, #4/3 pi r^3#. The cylinder has volume #pi r^2 l#.

Then we write,

#V = 4/3 pi r^3 + pi r^2 l#,
#l = V/(pi r^2) - 4/3 r#.

The surface area of the geometric object will be the surface area of a sphere with radius #r#, #4 pi r^2#, plus the curved surface area of a cylinder with radius #r# and length #l#, #2 pi r l#.

So we write,

#S=2 pi r l + 4 pi r^2#.

Substituting the definition of #l# in terms of #r# and #V# gives,

#S = 2V/r - 8/3 pi r^2 + 4 pi r^2#,
#S = 2Vr^(-1)+4/3pi r^2#.

Given that #V# is a fixed positive number, this function is increasing as #r -> + infty# and goes to #+infty# as #r -> 0#. Then it turns and obtains a minimum value as required in this interval.

We solve for the turning points by differentiating and equating with zero to find the value(s) of #r# for which the function turns.

#("d"S)/("d"r) = 8/3 pi r - 2V/r^2#,

Setting #("d"S)/("d"r)=0# and solving for #r# gives #r^3 = (3V)/(4pi)# which has one real positive root for #r#, #r = ((3V)/(4pi))^(1/3)#.