A container, opened from the top and made up of a metal sheet,

Solution:

We have $: r_{1}=20 \mathrm{~cm}$,

$\mathrm{r}_{2}=8 \mathrm{~cm}$ and $\mathrm{h}=16 \mathrm{~cm}$

$\therefore$ Volume of the frustum

$=\frac{\mathbf{1}}{\mathbf{3}} \pi h\left[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right]$

$=\frac{\mathbf{1}}{\mathbf{3}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 16\left[20^{2}+8^{2}+20 \times 8\right] \mathrm{cm}^{3}$

$=\frac{\mathbf{1}}{\mathbf{3}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 16 \times[400+64+160] \mathrm{cm}^{3}$

$=\frac{\mathbf{1}}{\mathbf{3}} \times \frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 16 \times 624 \mathrm{~cm}^{3}=\left[\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times \mathbf{1 6} \times \mathbf{2 0 3}\right] \mathrm{cm}^{3}$

$=\left[\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times \mathbf{1 6} \times \mathbf{2 0 3}\right] \div 1000$ litres

$=\frac{\mathbf{3 1 4} \times \mathbf{1 6} \times \mathbf{2 0 8}}{\mathbf{1 0 0 0 0 0}}$ litres

$\therefore$ Cost of milk $=` 20 \times \frac{\mathbf{3 1 4} \times \mathbf{1 6} \times \mathbf{2 0 8}}{\mathbf{1 0 0 0 0 0}}$ litres

$=` 208.998 \approx 209$

Now, slant height of the given frustum

$\ell=\sqrt{\mathbf{h}^{2}+\left(\mathbf{r}-\mathbf{r}_{2}\right)^{2}}=\sqrt{\mathbf{1 6}^{2}+(\mathbf{2 0}-\mathbf{8})^{2}}$

$=\sqrt{\mathbf{1 6}^{2}+12^{2}}=\sqrt{\mathbf{2 5 6}+144}=\sqrt{\mathbf{4 0 0}}=20 \mathrm{~cm}$

$\therefore$ Curved surface area $=\pi\left(r_{1}+r_{2}\right) \ell$

$=\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}}(20+8) \times 20 \mathrm{~cm}^{2}$

$=\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 28 \times 20 \mathrm{~cm}^{2}=1758.4 \mathrm{~cm}^{2}$

Area of the bottom $=\pi r^{2}$

$=\frac{\mathbf{3 1 4}}{\mathbf{1 0 0}} \times 8 \times 8 \mathrm{~cm}^{2}=200.96 \mathrm{~cm}^{2}$

$\therefore$ Total area of metal required

$=1758.4 \mathrm{~cm}^{2}+200.96 \mathrm{~cm}^{2}=1959.36 \mathrm{~cm}^{2}$

Cost of metal required $=\frac{\mathbf{8}}{\mathbf{1 0 0}} \times 1959.36 \mathrm{~cm}^{2}$

$=156.75$