A container, opened from 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 the milk which can completely fill the container, at the rate of Rs.20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs.8 per 100 cm^{2}. [Take π = 3.14]

#### Solution

Radius (*r*_{1}) of upper end of container = 20 cm

Radius (*r*_{2}) of lower end of container = 8 cm

Height (*h*) of container = 16 cm

Slant height (*l*) of frustum = `sqrt((r_1-r_2)^2+h^2)`

`=sqrt((20-8)^2+(16)^2)`

`=sqrt((12)^2+(16)^2) = sqrt(144+256)`

= 20 cm

Capacity of container = Volume of frustum

`=1/3pih(r_1^2+r_2^2+r_1r_2)`

`=1/3xx3.14xx16xx[(20)^2+(8)^2+(20)(8)]`

`=1/3xx3.14xx16(400+64+160)`

`=1/3xx3.14xx16xx624`

= 10449.92 cm^{3}

= 10.45 litres.

Cost of 1 litre milk = Rs 20

Cost of 10.45 litre milk = 10.45 × 20

= Rs 209

Area of metal sheet used to make the container

`=pi(r_1+r_2)l + pir_2^2`

= π (20 + 8) 20 + π (8)^{2}

= 560 π + 64 π = 624 π cm^{2}

Cost of 100 cm^{2} metal sheet = Rs 8

Cost of 624 π cm^{2 }metal sheet = `(624 xx 3.14 xx 8)/100`

= 156.75

Therefore, the cost of the milk which can completely fill the container is

Rs 209 and the cost of metal sheet used to make the container is Rs 156.75.