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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 the milk which can completely fill the container

4. 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 $\pi = 3.14$)

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Firstly we will calculate the slant height of the cone :

$l^2\ =\ \left ( r_1^2\ -\ r^2_2 \right )\ +\ h^2$

$l^2=\ \left ( 20^2\ -\ 8^2 \right )\ +\ 16^2$

$l\ =\ 20\ cm$

Now, the volume of the frustum is :

$=\ \frac{1}{3}\pi h\left ( r^2_1\ +\ r^2_2\ +\ r_1r_2 \right )$

$=\ \frac{1}{3}\pi \times 16\left ( 20^2\ +\ 8^2\ +\ 20\times 8 \right )$

$=\ 10449.92\ cm^3$

=   Capacity of the container.

Now, the cost of 1-litre milk is  Rs. 20.

Then the cost of 10.449-litre milk will be   $=\ 10.45\times 20\ =\ Rs.\ 209$

The metal sheet required for the container is  :

$=\ \pi (r_1\ +\ r_2)l\ +\ \pi r_2^2$

$=\ \pi (20\ +\ 8)20\ +\ \pi\times 8^2$

$=\ 624 \pi\ cm^2$

Thus cost for metal sheet is

$=\ 624 \pi\ \times \frac{8}{100}$

$=\ Rs.\ 156.57$

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