Question

A container opened at the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container, at the rate of ₹ 50 per litre. Also find the cost of metal sheet used to make the container, if it costs ₹ 10 per 100 cm². (Take π = 3⋅14)

Answer 1

We have to find the cost of milk which can completely fill the container
The volume of container = Volume of a frustum
= `1/3`πh(r₁² + r₂² + r₁r₂ )

Here,
height = 16 cm
radius of upper end = 20 cm
And radius of lower end = 8 cm
Plugging the values in the formula we get

Volume of container = `1/3` x 3.14 x 16 ((20)² + (8)² + 20 x 8)
= `1/3` x 50.24 (400 + 64 + 160 )
= `1/3` x 50.24 (624)
= 10449.92 cm³        
=10.449 litre                 ....(∵ 1 litre=1000 cm³)

Cost of 1-litre milk is Rs 50
Cost of 10.449 litre milk = 50 x 10.449 = Rs 522.45                                 
We will find the cost of metal sheet to make the container
Firstly, we will find the area of container

Area of container = Curved surface area of the frustum + area of bottom circle    ...(∵ container is closed from bottom)

Area of container = π(r₁ + r₂)l + πr²
Now, we will find l
l = `sqrt(h^2 + (r_1 − r_2)^2)`

`sqrtl = sqrt((16)^2+(20−8)^2)`

`sqrtl = sqrt((16)^2+(12)^2)`

`sqrtl = sqrt( 256 + 144)`

`sqrtl = sqrt(400)`
`l = 20 cm`
              
Area of frustum = 3.14 x 20(20 + 8)                   
                          = 1758.4 cm²

Area of bottom circle = 3.14 x 8² = 200.96 cm²

Area of container = 1758.4 + 200.96  =1959.36 cm²

Cost of making 100 cm² = Rs 10
Cost of making 1 cm² = `10/100` = Rs. 110
Cost of making 1959.36 cm² = `1/10` x 1959.36 = 195.936                       
Hence, cost of milk is Rs 522.45
And cost of metal sheet is Rs 195.936.