$A$ 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 the metal sheet used to make the container,if it costs $Rs. \, 8$ per $100 \, cm^2$. (Take $\pi = 3.14$)

(N/A) 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 $= \sqrt{(r_1 - r_2)^2 + h^2} = \sqrt{(20 - 8)^2 + 16^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, cm$
Capacity of container $=$ Volume of frustum $= \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$
$= \frac{1}{3} \times 3.14 \times 16 \times (20^2 + 8^2 + 20 \times 8) = \frac{1}{3} \times 3.14 \times 16 \times (400 + 64 + 160) = \frac{1}{3} \times 3.14 \times 16 \times 624 = 10449.92 \, cm^3 = 10.44992 \, litres \approx 10.45 \, litres$
Cost of $1 \, litre$ milk $= Rs. \, 20$
Cost of $10.45 \, litres$ milk $= 10.45 \times 20 = Rs. \, 209$
Area of metal sheet used $= \pi (r_1 + r_2) l + \pi r_2^2 = 3.14 \times (20 + 8) \times 20 + 3.14 \times 8^2 = 3.14 \times 28 \times 20 + 3.14 \times 64 = 1758.4 + 200.96 = 1959.36 \, cm^2$
Cost of $100 \, cm^2$ metal sheet $= Rs. \, 8$
Cost of $1959.36 \, cm^2$ metal sheet $= \frac{1959.36 \times 8}{100} = Rs. \, 156.7488 \approx Rs. \, 156.75$
Therefore,the cost of the milk is $Rs. \, 209$ and the cost of the metal sheet is $Rs. \, 156.75$.