Question

Sumit has bought a closed cylindrical dustbin. The radius of the dustbin is ‘r' cm and height is 'h’ cm. It has a volume of 20π cm³.

1. Express ‘h’ in terms of ‘r’, using the given volume.
2. Prove that the total surface area of the dustbin is `2πr^2 + (40π)/r`
3. Sumit wants to paint the dustbin. The cost of painting the base and top of the dustbin is ₹ 2 per cm² and the cost of painting the curved side is ₹ 25 per cm². Find the total cost in terms of ‘r’, for painting the outer surface of the dustbin including the base and top.
4. Calculate the minimum cost for painting the dustbin.

Answer 1

Given, radius of dustbin is r and height is h.

a. Volume of dustbin,

V = 20π cm³

Then, πr²h = 20π

`\implies` r²h = 20 

`\implies` h = `20/r^2`

b. T.S.A. of dustbin = C.S.A + 2 base area

= 2πrh + 2πr²

= `2πr xx 20/r^2 + 2πr^2`

= `(40π)/r + 2πr^2`

= `2πr^2 + (40π)/r`

Hence proved

c. C.S.A. of dustbin = `(40π)/r cm^2`

Then, cost of painting (SA) = `(40π)/r xx 25`

= ₹ `(1000π)/r`

Base and top area of dustbin = 2πr²

Then, cost of painting (top and bottom)

= 2πr² × 2

= ₹ 4πr²

Total cost of painting = ₹ `((1000π)/r + 4πr^2) `

d. Cost of painting,

C = `(1000π)/r + 4πr^2`

∴ `(dC)/(dr) = -(1000π)/r^2 + 8πr`

For minimum cost,

Put `(dC)/(dr)` = 0

`-(1000π)/r^2 + 8πr` = 0

`\implies` 8πr = `(1000π)/r^2`

`\implies` r³ = 125

`\implies` r = 5 cm

And `(d^2C)/(dr^2) = (2000π)/r^3 + 8π > 0`

Then, `(d^2C)/(dr^2)` is minimum for r = 5 cm.

Then, minimum cost of painting = `(1000π)/5 + 8π(5)`

= 200π + 40π

= ₹ 240π