A milk container is made of metal sheet in the shape of a frustum of a cone whose volume is $\displaystyle 10459\frac{3}{7} \ cm^{3}$.The radii of its lower and upper circular ends are$\displaystyle \ 8$ cm and $\displaystyle 20$ cm respectively. find the cost of metal sheet used in making the container at rate Rs. 1.40 per square centimeter.

Given: Volume of the container $( V)=10459\frac{3}{7}$ , raddi of upper circular end $( r)=8\ cm$, raddi of lower circular end $( R)=20\ cm$ and metal sheet cost$=1.40$  per square feet.

To do: To find total cost of metal sheet to be used to in making the container.

Solution:
Radii of the lower end $r=8\ cm$

Radii of the upper end $R=20\ cm$

Let we assume the height of the frustum $h\ cm$.

Then volume of the frustum V$=\frac{\pi }{3} h\left( R^{2} +r^{2} +Rr\right)$

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

$=\frac{22}{7} \times \frac{1}{3} \times 624\times h$

$=\frac{22}{7} \times h\times 208$

Here volume is given V$=10459\frac{3}{7}$

Substituting its value,

$\frac{73216}{7} =\frac{22}{7} \times h\times 208$

$\Rightarrow h=\frac{73216}{22\times 208}$

$\Rightarrow h=16\ cm$

Total surface area of the container A
$=\pi( R+r)\sqrt{\left(( R-r)^{2} +h^{2} \ \right)} +πr^{2}$

$=\ \ \frac{22}{7}( 20+8)\sqrt{\left(( 20-8)^{2} +( 16)^{2} \ \right)} +\frac{22}{7} \times 8\times 8$

$=\frac{22}{7}( 28)\sqrt{\left(( 12)^{2} +( 16)^{2} \ \right)} +\frac{22}{7} \times 8\times 8$

$=\frac{22}{7}\left( 28\sqrt{( 144+256)} +64\right)$

$=\frac{22}{7}\left( 28\times \sqrt{400} +64\right)$

$=\frac{22}{7}( 28\times 20+64)$

$=\frac{22}{7} \times 624$

Here metal cost is given 1.40 per cm square.

Total metal cost$=surface\ area\times 1.40$

$=\frac{22}{7} \times 624\times 1.40$

$=2745.60$.

Thus cost of metal sheet is Rs. 2745.60 in making the container.
 

Updated on: 10-Oct-2022

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