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# 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 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. (Take $$\pi \ = \ 3.14$$ ).

Here R = 20 cm, r = 8 cm and h = 16 cm

Capacity of the container = Volume of the frustum $$= \ \frac{1}{3} \pi h(R^2 \ + \ r^2 \ + \ Rr)$$

$$= \ \frac{1}{3} \ × \ 3.14 \ × \ 16(20^2 \ + \ 8^2 \ + \ 20 \ × \ 8) \$$

$$= \ \frac{50.24}{3} \ × \ (400 \ + \ 64 \ + \ 160)$$

$$= \ \frac{50.24}{3} \ × \ 624 \$$

$$= \ 50.24 \ × \ 208$$

$$= \ \frac{10449.92}{1000}$$ litres

Cost of milk @ Rs 20 per litre $$= \ 20 \ × \ \frac{10449.92}{1000}$$

$$= \ 208.99$$ $$\approx$$ Rs $$209$$

To find the slant height

$$l \ = \ \sqrt{16^2 \ + \ 12^2} \$$
$$= \ \sqrt{256 \ + \ 144}$$
$$= \ \sqrt{400} \$$
$$= \ 20$$ cm

Curved Surface Area of the container = $$\pi (R \ + \ r)l$$

$$= \ \frac{22}{7} \ × \ (20 \ + \ 8) \ × \ 20$$

$$= \ \frac{22}{7} \ × \ 28 \ × \ 20 \ = \ 1758.4$$ cm2

Area of the bottom of the container = $$\pi r^2$$

$$= \ \frac{22}{7} \ × \ (8)^2 \ = \ \frac{22}{7} \ × \ 64 \ = \ 200.96$$ cm2

$$\therefore$$ Total area of metal required $$= 1758.4 \ + \ 200.96 \ = \ 1959.36$$ cm2

Cost of metal sheet used to manufacture the container @ Rs 8 per 100 cm2 $$= \ \frac{8}{100} \ × \ 1959.36 \ =$$ Rs $$156.75$$