Question

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]
​

Answer 1

Answer:

the capacity is 5 litre

hope it helps you

mark as brainliest

Answer 2

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

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

$\bf \: ⟼ h = 2r$

$\bf \: ⟼ \dfrac{h}{r} = 2$

$\bf \: ⟼ h : r = 2 : 1$