A milk container is made of metal sheet in the shape of frustum of a cone whose volume is

Solution:

We have,

Radius of the upper end, $R=20 \mathrm{~cm}$ and

Radius of the lower end, $r=8 \mathrm{~cm}$

Let the height of the container be $h$.

As,

Volume of the container $=10459 \frac{3}{7} \mathrm{~cm}^{3}$

$\Rightarrow \frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)=\frac{73216}{7}$

$\Rightarrow \frac{1}{3} \times \frac{22}{7} \times h \times\left(20^{2}+8^{2}+20 \times 8\right)=\frac{73216}{7}$

$\Rightarrow \frac{22 h}{21} \times(400+64+160)=\frac{73216}{7}$

$\Rightarrow \frac{22 h}{21} \times 624=\frac{73216}{7}$

$\Rightarrow h=\frac{73216 \times 21}{7 \times 22 \times 624}$

$\Rightarrow h=16 \mathrm{~cm}$

Also,

The slant height of the container, $l=\sqrt{(R-r)^{2}+h^{2}}$

$=\sqrt{(20-8)^{2}+16^{2}}$

$=\sqrt{12^{2}+16^{2}}$

$=\sqrt{144+256}$

$=\sqrt{400}$

$=20 \mathrm{~cm}$

Now,

Total surface area of the container $=\pi l(R+r)+\pi r^{2}$

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

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

$=\frac{22}{7} \times(560+64)$

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

$=\frac{13728}{7} \mathrm{~cm}^{2}$

So, the cost of metal sheet $=1.4 \times \frac{13728}{7}=₹ 2745.60$

Hence, the cost of the metal sheet used for making the milk container is ₹2745.60.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.