Math, asked by usamabin, 3 months ago


9. Find the dimensions of the closed cylindrical can that will have a capacity of k units of volume and will use the
minimum amount of material. Find the ratio of the height h to the radius r of the top and bottom. [h/r=2]

Answers

Answered by Aizaa9000
0

Answer:

the capacity is 5 litre

hope it helps you

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usamabin: step by step answer in this question plz
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usamabin: thanks and fastly
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Answered by mathdude500
1

Question :-

  • Find the dimensions of the closed cylindrical can that will have a capacity of k units of volume and will use the minimum amount of material. Find the ratio of the height h to the radius r of the top and bottom.

Answer

Given :-

  • Volume of cylinderical can = k

To find :-

  • Ratio of the height h to the radius r of the top and bottom.

\begin{gathered}\Large{\bold{\pink{\underline{Formula \:  Used \::}}}}  \end{gathered}

\bf \:  ⟼ \dfrac{d}{dx} {x}^{n}  =  {nx}^{n - 1}

\bf \:  ⟼ \dfrac{d}{dx}\dfrac{1}{x}  =  - \dfrac{1}{ {x}^{2} }

\bf \:  ⟼ \dfrac{d}{dx}(\dfrac{1}{ {x}^{2} } ) =  - \dfrac{2}{ {x}^{3} }

\begin{gathered}\Large{\bold{\pink{\underline{CaLcUlAtIoN\::}}}}\end{gathered}

\begin{gathered}\bf\red{Let}\end{gathered}

\begin{gathered}\longmapsto\:\:\bf{Radius\: \:is \:r\:units}. \\ \end{gathered}

\begin{gathered}\longmapsto\:\:\bf{Height \: \:is \:h\:units}. \\ \end{gathered}

{{ \boxed{\large{\bold\green{Volume_{(Cylinder)}\: = \:\pi r^2 h }}}}}

\bf \:  ⟼ k = \pi \:  {r}^{2} h

\bf \:  ⟼ h = \dfrac{k}{ {\pi \: r}^{2} }  \:⟼(1)

{{ \boxed{\small{\bold\green{Total \:  Surface \:  Area_{(Cylinder)}\: = \:2\pi r(r + h )}}}}}

Now, Cost C, of construction the can is minimum, when surface area is minimum.

\begin{gathered}\bf\red{So,}\end{gathered}

\bf \:  ⟼ C = 2\pi \: r(h + r)

On substituting the value of h, from equation (1), we get

\bf \:  ⟼ C = 2\pi \: r(\dfrac{k}{ {\pi \: r}^{2} } + r)

\bf \:  ⟼ C = 2\pi \:(\dfrac{k}{ {\pi \: r}} +  {r}^{2} )

Differentiate w. r. t. r both sides, we get

\bf \:  ⟼ \dfrac{dC}{dr} = 2\pi \: ( - \dfrac{k}{ {\pi \: r}^{2} } + 2r) \: ⟼ \: (2)

For maximum or minimum value

\bf \:  ⟼ \dfrac{dC}{dr} = 0

\bf \:  ⟼ 2\pi \: ( - \dfrac{k}{ {\pi \: r}^{2} } + 2r)

\bf \:  ⟼   \dfrac{k}{ {\pi \: r}^{2} }  =  2r

\bf \:  ⟼ k = 2\pi \:  {r}^{3}  \: ⟼ \: (3)

Differentiate equation(2) w. r. t. r, we get

\bf \:  ⟼ \dfrac{ {d}^{2} C}{d {r}^{2} } = 2\pi \: ( \dfrac{2k}{ {\pi \: r}^{3} } + 2)  > 0

\bf \:  ⟼    C \: is \: minimum.

On substituting equation (3) in equation (1), we get

\bf \:  ⟼ h = \dfrac{2 \: \pi \:  {r}^{3} }{\pi \:  {r}^{2} }

\bf \:  ⟼ h = 2r

\bf \:  ⟼ \dfrac{h}{r}  = 2

\bf \:  ⟼ h : r = 2 : 1


usamabin: Thanks bro
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