 # 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 cm2.

Given

Height of frustum (h)= 16 cm

Let radius of smaller circle end (r) = 8 cm

Let radius of larger circle end (R) = 20 cm

Find out

We have to find the cost of metal sheet used to make the container

Solution

The amount of milk to fill the container = Volume of frustum

= 1/3 πh (r2 + R2 + rR)

= 1/3 × 3.14 × 16 (8×8 + 20×20 + 8×20) (Taking π=3.14)

= 1/3 × 3.14 × 16 × (624)

= 10449.92 cm3

Cost of 1 litre milk = ₹ 20

We know that

1 m= 1000 cm= 1 litre

10449.92 cm= (1/1000) ×10449.92 litres

=10.4492

cost of 10449.92 cm= 10.4492 × 20

= ₹ 208.998

Now, metal sheet used to make the container = Curve surface area + area of lower base

= (π(r+R)l) + (πr2)

= π ((r+R)l + r2)

We know that

Slant height (l) = √(h2+(R-r)2)

= π ((20+8) × (√(162+(20-8)2)) + (8 × 8))

On Taking π value = 3.14

= 3.14 × (28 × √400 + 64)

= 3.14 × (624)

= 1959.36 cm2

Hence, the metal sheet used to make the container = 1959.36 cm2

The cost of 100 cm2= ₹ 8

1959.36 cm2 = (8/100) × 1959.36

= ₹156.748

Hence, the cost of the milk which can completely fill the container = ₹ 208.998

and, the cost of the metal sheet used to make the container = ₹156.748