# A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm

Question:

A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20cm respectively. Find
(i) the cost of metal sheet used to make the container if it costs ₹ 10 per 100 cm2

(ii) the cost of milk at the rate of ₹ 35 per litre which can fill it completely.

Solution:

Let $r=8 \mathrm{~cm}, R=20 \mathrm{~cm}, h=16 \mathrm{~cm}$.

$\Rightarrow l=\sqrt{(R-r)^{2}+h^{2}}=\sqrt{(20-8)^{2}+16^{2}}=\sqrt{144+256}=\sqrt{400}$

$\Rightarrow l=20 \mathrm{~cm}$

(i)

Total metal sheet required to make the container = Curved surface area of frustum + Area of the base

$=\pi(r+R) l+\pi r^{2}$

$=\pi(8+20) 20+\pi(8)^{2}=560 \pi+64 \pi=624 \pi=1961.14 \mathrm{~cm}^{2}$

The cost for 100 cm2 of sheet = Rs. 10.

The cost of $1 \mathrm{~cm}^{2}$ of sheet $=\frac{10}{100}=$ Rs. $0.1$.

The cost of $1961.14 \mathrm{~cm}^{2}$ of sheet $=1961.14 \times 0.1=$ Rs. $196.11$

DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =

$\frac{\pi h}{3}\left[r^{2}+R^{2}+r R\right]=\frac{22}{7} \times \frac{16}{3}\left[8^{2}+20^{2}+160\right]=\frac{22}{7} \times \frac{16}{3}[64+400+160]$

$=\frac{22}{7} \times \frac{16}{3} \times 624=\frac{73216}{7} \mathrm{~cm}^{3} .$

we know that 1 l=1000 cm3 or 1 cm3 = 0.001 l.

$\Rightarrow$ Volume $=\frac{73216}{7} \times 0.001=\frac{73216}{7} \times \frac{1}{1000}=\frac{73.216}{7} l$

Cost of milk is Rs. 35 per litre.

Hence, the cost at which this frustum can be filled $=\frac{73.216}{7} \times 35=$ Rs $366.08$.