A container, open at the top, made up of metal sheet is in the form of the frustum of a cone of height 16 cm with radii of lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of < 40 per litre and also find the cost of metal sheet used if it costs < 10 per 100 cm2. (Use π = 3.14)
We have to find cost of milk which can completely fill container, for that we need to find volume (in litres)
Given:
Radius of the upper end of the container = r1 = 20 cm
Radius of the lower end of the container = r2 = 8 cm
Height of the container = 16 cm
Slant height, l = √{h2 + (r1 – r2)2}
= √{(16)2 +(20 – 8)2
= √256 + (12)2
= √256 + 144
= √400
= 20 cm
Now,
= 10445.76 cm3
= 10.44576 Litre
A litre of milk cost Rs 40
So, total cost of filling the container with milk = Rs 40 × 10.44576
= Rs 417.83
Now, we need to find cost of metal, for that we need to find the area of container
Since container is closed from bottom,
Surface Area of the container
= CSA of the frustum + Area of circular base
= πl(r1+r2) + πr22
= 3.14 × [{20 × (20 + 8)} + (8)2]
= 3.14 × [400 + 160 + 64]
= 3.14 × 624
= 1959.36 cm2
Cost of 100cm2 of metal sheet = Rs 10
= Rs 195.93
Hence,
Cost of milk = Rs 417.83
Cost of metal sheet = Rs 195.93