Question

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the cost to make the can if the metal for the sides will cost $1.25 per 2 cm and the metal for the bottom will cost $2.00 per 2 cm ?

Answer 1

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V=
`\pi r^2h`

According to the problem,

`\pi r^2 h=25`

`\Rightarrow h=(25)/(\pi r^2)`

The surface area of the base of the can is =
`\pi r^2`

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00×
`\pi r^2`)

The lateral surface area of the can is =
`2\pi rh`

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25×
`2\pi rh`)

`=\$2.5 \pi r h`

Total cost of metal is C= 2.00
`\pi r^2`+
`2.5 \pi r h`

Putting
`h=(25)/(\pi r^2)`

`\therefore C=2\pi r^2+2.5 \pi r * (25)/(\pi r^2)`

`\Rightarrow C=2\pi r^2+ (62.5)/( r)`

Differentiating with respect to r

`C'=4\pi r- (62.5)/( r^2)`

Again differentiating with respect to r

`C''=4\pi + (125)/( r^3)`

To find the minimize cost, we set C'=0

`4\pi r- (62.5)/( r^2)=0`

`\Rightarrow 4\pi r=(62.5)/( r^2)`

`\Rightarrow r^3=(62.5)/( 4\pi)`

⇒r=1.71

Now,

`\left C''\right|_(x=1.71)=4\pi +(125)/(1.71^3)>0`

When r=1.71 cm, the metal cost will be minimum.

Therefore,

`h=(25)/(\pi* 1.71^2)`

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.