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 $Rs.8per100c{m}^{2}$.

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Solution

It is given that,$R=20cm\phantom{\rule{0ex}{0ex}}r=8cm\phantom{\rule{0ex}{0ex}}h=16cm$Step 1 - Finding the cost of milk that it can contain.$\therefore Volumeoffrustum=\frac{1}{3}\mathrm{\pi h}\left({R}^{2}+{r}^{2}+Rr\right)\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=\frac{1}{3}×3.14×16\left({20}^{2}+{8}^{2}+20×8\right)\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=\frac{1}{3}×3.14×16\left(624\right)\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=10449.92{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}$We know that $1c{m}^{3}=1ml$$10449.92c{m}^{3}=1×10499.92=10499.92ml$We also know that, $1000ml=1litre$Therefore, $10449.92ml=\frac{10499.92}{1000}litre=10.49992litre$$\therefore \mathrm{Cost}\mathrm{of}1\mathrm{litre}\mathrm{milk}=₹20\phantom{\rule{0ex}{0ex}}\mathrm{So}\mathrm{Cost}\mathrm{of}10499.92\mathrm{litre}\mathrm{milk}=20×10499.92=208.998$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 containerWe know the relation between the slant height, radii and height of the frustum is${l}^{2}={h}^{2}+{\left(R-r\right)}^{2}\phantom{\rule{0ex}{0ex}}{l}^{2}={16}^{2}+{\left(20-8\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{l}^{2}=256+144\phantom{\rule{0ex}{0ex}}l=\sqrt{400}\phantom{\rule{0ex}{0ex}}l=20cm$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 = $\frac{8×1959.36}{100}=156.748$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 $₹156.748$.

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