A tank is made in the shape of a closed cylinder with a hemisphere at one end. The tank has a volume of 30 m3

Question

A tank is made in the shape of a closed cylinder with a hemisphere at one end. The

tank has a volume of 30 m3. Let r be the radius of the cylinder (and hemisphere),

and h be the height of the cylinder.

(a) Use the surface area formulae below to express the surface area of the can in terms of r and h.

(b) Use the volume formulae below to express the volume of the can in terms of r and h.

(c) Using the fact the volume is fixed, rearrange (b) to write h in terms of r.

(d) Substitute your expression for the h in (c) into your surface area expression

from (a). Afterwards, r should be the only variable in your surface area expression.

(e) Use calculus to find the value of r where the surface area is a minimum.

(f) What is the value of h at the minimum? Can you make a general statement

about the dimensions of any tank comprised of a cylinder and a hemisphere?

Useful formulae:

volume of a sphere =4πr3/3

volume of a cylinder = πr^2h

Surface area of a sphere = 4πr2

Surface area of an open cylinder = 2πrh

Answer 1

Hi Ramon,

First, have a visualization of the closed cyllinder with the hemisphere - then you can see how these surface areas and volumes add up intuitively.

a) Surface area of can = 1/2 * S_sphere + S_{open cyllinder} + S_circle = 2 π r² + 2 π r *h + π r² = 3π r² + 2 π r *h

b) Volume of can =V = 1/2 * V_sphere + V _cyllinder = 2 π r³/3 + π r²*h

c) π r²*h = V - 2 π r³/3 , thus: h = V/(π r²) - 2r/3

d) S = 3π r² + 2 π r *h = 3π r² + 2 π r * (V/(π r²) - 2r/3) = 3π r² + 2V/r - 4π r²/3 = 5π r²/3 + 2V/r

e) To find the extreme points (min or max), you take the derivative and set it to 0. To distinguish if it is a min or a max, you have to look at the second derivative - your point is a min if the second derivative is > 0 at that point and max if it is <0.

dS/dr = 10π r/3 - 2V/r²

And you can finish it from here. Lemme know if you have questions.