A cylindrical can is to be made to hold 1 liter of oil. Find the radius that will minimize the cost of the metal to manufacture the can.

For minimum amount of material
$h = 2r$
 

 

$V = \pi r^2 h$

$1000 = \pi r^2 (2r)$

$1000 = 2\pi r^3$

$r = 5.42 ~ \text{cm}$

Capacity of the can:
$V = \pi r^2 h$

$1000 = \pi r^2 h$

$h = \dfrac{1000}{\pi r^2}$
 

 

Area of metal to manufacture the can
$A = 2A_b + A_L$

$A = 2\pi r^2 + 2\pi rh$

$A = 2\pi r^2 + 2\pi r \left( \dfrac{1000}{\pi r^2} \right)$

$A = 2\pi r^2 + \dfrac{2000}{r}$
 

To minimize the cost of the metal:
$\dfrac{dA}{dr} = 4\pi r - \dfrac{2000}{r^2} = 0$

$4\pi r = \dfrac{2000}{r^2}$

$r^3 = \dfrac{2000}{4\pi}$

$r = 5.42 ~ \text{cm}$