Question

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 $16cm$ with radii of its lower and upper ends as $8cm$ and $20cm$, 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 $Rs.8per100c{m}^2$.

Answer 1

It is given that,

$$\begin{gathered} R=20cm \\ r=8cm \\ h=16cm \end{gathered}$$

Step 1 - Finding the cost of milk that it can contain.

$$\begin{gathered} \therefore Volumeoffrustum=\frac{1}{3}\mathrm{\pi h}\left(R^2+r^2+Rr\right) \\ \mathrm{Volume}\mathrm{of}\mathrm{frustum}=\frac{1}{3}\times 3.14\times 16\left({20}^2+8^2+20\times 8\right) \\ \mathrm{Volume}\mathrm{of}\mathrm{frustum}=\frac{1}{3}\times 3.14\times 16\left(624\right) \\ \mathrm{Volume}\mathrm{of}\mathrm{frustum}=10449.92{\mathrm{cm}}^3 \end{gathered}$$

We know that $1c{m}^3=1ml$

$10449.92c{m}^3=1\times 10499.92=10499.92ml$

We also know that, $1000ml=1litre$

Therefore, $10449.92ml=\frac{10499.92}{1000}litre=10.49992litre$

$$\begin{gathered} \therefore \mathrm{Cost}\mathrm{of}1\mathrm{litre}\mathrm{milk}=₹20 \\ \mathrm{So}\mathrm{Cost}\mathrm{of}10499.92\mathrm{litre}\mathrm{milk}=20\times 10499.92=208.998 \end{gathered}$$

Hence, the cost of milk in the container is $₹208.992$.

Step 2 - Finding the cost of the metal sheet required to make the container.

Since, the container is open from the top.

So, the surface area of the container = Curved surface area of container + Area of the base of the container

We know the relation between the slant height, radii and height of the frustum is

$$\begin{gathered} l^2=h^2+{\left(R-r\right)}^2 \\ {l}^2={16}^2+{\left(20-8\right)}^2 \\ {l}^2=256+144 \\ l=\sqrt{400} \\ l=20cm \end{gathered}$$

Now,

$\begin{array}{rcl}Surfaceareaofthecontainer& =& Curvedsurfaceareaoffrustum+Areaofthebaseoffrustum\\ Surfaceareaofthecontainer& =& \mathrm{\pi }\left(\mathrm{R}+\mathrm{r}\right)\mathrm{l}+{\mathrm{\pi R}}^2\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}& =& \mathrm{\pi }\left\{\left(\mathrm{R}+\mathrm{r}\right)\mathrm{l}+{\mathrm{R}}^2\right\}\\ & & \\ \mathrm{Taking}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{\pi }& =& 3.14\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}& =& 3.14\left\{\left(20+8\right)20+8^2\right\}\\ \mathrm{Surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{container}& =& 1959.36{\mathrm{cm}}^2\end{array}$

Now, the metal sheet used to make the container is $1959.36c{m}^2$.

The cost of $100c{m}^2$ of the sheet = $₹8$

The cost of $1959.36c{m}^2$ of the sheet = $\frac{8\times 1959.36}{100}=156.748$

Therefore, the cost of the metal sheet used to make the container is $₹156.748$.

Hence, the cost of milk in the container is $₹208.992$ and the cost of the metal sheet used to make the container is $₹156.748$.