Question

A person has manufactured a water tank in the shape of a closed right circular cylinder. The volume of the cylinder is `539/2` cubic units. If the height and radius of the cylinder are h and r.

1. Express h in terms of radius r and given volume. 
2. Let the total surface area of the closed cylinder tank be S; express S in terms of radius r. 
3. If the total surface area of the tank is minimum, then prove that radius r = `7/2` units.
4. Find the height of the tank.

Answer 1

Given: The tank is a closed right circular cylinder.

Volume = `539/2` cubic units

Height = h, radius = r

a. Express h in terms of radius and given volume.

Volume (V) = πr²h

`539/(2) = πr^2h`

h = `(539/2)/(πr^2)`

h = `539/(2πr^2)`

b. Let the total surface area of the closed cylinder tank be S.

Expressing S in term of radius r.

S = 2πrh + 2πr²

S = `2πr 539/(2πr^2) + 2πr^2`

S = `2πr^2 + 539/r`

 c. `(ds)/(dr) = 4πr - 539/r^2`

Setting, `(ds)/(dr)` = 0 for stationary point, we get

`4πr - 539/r^2 = 0`

4πr = `539/r^2`

4πr³ = 539

`r^3 = 539/(4π)`

`r^3 = 539/(4 xx 22/7)`

`r^3 = 539/(88/7)`

`r^3 = 539 xx 7/88`

`r^3 = (49 xx 11 xx 7)/(8 xx 11)`

`r^3 = 343/8`

`r^3 = (7/2)^3`

r = `7/2` units

Now, `(d^2s)/(dr^2) = 4π + (539 xx 2)/r^3 > 0`, when r = `7/2` units.

Therefore, the total surface area (S) of the tank is minimum when r = `7/2` units.

d. Finding the height of the tank:

h = `539/(2πr^2)`

h = `(539 xx 4 xx 7)/(2 xx 22 xx 7 xx 7)`

h = `(49 xx 11 xx 7 xx 4)/(2 xx 22 xx 7 xx 7)`

h = 7 units