Question

Prove that volume of closed cylinder of given total surface area is maximum when height is equals to diameter of base.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

Height of cylinder be h units

Radius of cylinder be r units

S represents the total surface area of cylinder

V represents Volume of cylinder.

We know, Total Surface Area (S) of cylinder of radius r and height h is given by

$\rm \: S = 2\pi \: r(h + r)$

$\rm \: S = 2\pi \: rh + 2\pi \: {r}^{2}$

$\rm\implies \:h = \dfrac{S - {2\pi \: r}^{2} }{2\pi \: r} - - - (1)$

Now, we know Volume of cylinder (V) of radius r and height h is given by

$\rm \: V = \pi \: {r}^{2} h$

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

$\rm \: V = \pi \: {r}^{2} \bigg(\dfrac{S - {2\pi \: r}^{2} }{2\pi \: r} \bigg)$

$\rm \: V =r \bigg(\dfrac{S - {2\pi \: r}^{2} }{2} \bigg)$

$\rm \: V =\dfrac{Sr - {2\pi \: r}^{3} }{2}$

On differentiating both sides w. r. t. r, we get

$\rm \: \dfrac{d}{dr}V =\dfrac{d}{dr}\dfrac{Sr - {2\pi \: r}^{3} }{2}$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

So, using this result, we get

$\rm \: \dfrac{dV}{dr} =\dfrac{S - {6\pi \: r}^{2} }{2} - - - - (2)$

For maxima or minima, we substitute

$\rm \: \dfrac{dV}{dr} = 0$

$\rm \: \dfrac{S - {6\pi \: r}^{2} }{2} = 0$

$\rm\implies \:S = 6\pi \: {r}^{2} - - - - (3)$

Now, From equation (2), we have

$\rm \: \dfrac{dV}{dr} =\dfrac{S - {6\pi \: r}^{2} }{2}$

On differentiating both sides w. r. t. r, we get

$\rm \: \dfrac{ {d}^{2} V}{d {r}^{2} } =\dfrac{d}{dr}\bigg(\dfrac{S - {6\pi \: r}^{2} }{2}\bigg)$

$\rm \: \dfrac{ {d}^{2} V}{d {r}^{2} } =\dfrac{0 - {12\pi \: r}^{} }{2}$

$\rm\implies \:\rm \: \dfrac{ {d}^{2} V}{d {r}^{2} } = - 6\pi \: r < 0$

$\rm\implies \:V \: is \: maximum$

Now, Substitute the value of S from equation (3) in equation (1), we get

$\rm \: \:h = \dfrac{6\pi \: {r}^{2} - {2\pi \: r}^{2} }{2\pi \: r}$

$\rm \: \:h = \dfrac{4\pi \: {r}^{2}}{2\pi \: r}$

$\rm\implies \:h = 2r$

Hence, Proved

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Basic Concept Used :-

Let y = f(x) be a given function.

To find the maximum and minimum value, the following steps are follow :

1. Differentiate the given function.

2. For maxima or minima, put f'(x) = 0 and find critical points.

3. Then find the second derivative, i.e. f''(x).

4. Apply the critical points ( evaluated in second step ) in the second derivative.

5. Condition :-

The function f (x) is maximum when f''(x) < 0.

The function f (x) is minimum when f''(x) > 0.

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Question:-

Prove that volume of closed cylinder of given total surface area is maximum when height is equals to diameter of base.

Given:-

• A right circular cylinder of given surface and maximum volume.

To Prove:-

• Height of cylinder is equal to diameter of base.

Solution:-

• Let, radius of the cylinder = r.

• Height of the cylinder = h.

Surface area:

${ \boxed{ \sf \large \red{S = 2\pi rh + 2\pi r {}^{2} }}} \: \: \: ...(1)$

$\sf \large⇒  \frac{S - 2\pi r {}^{2} }{2\pi r} = h$

Volume of cylinder is:

${ \boxed{ \sf \large \red{V = \pi r {}^{2}h }}}$

$\sf \large⇒ v = \pi r {}^{2} \bigg ( \frac{S - 2\pi r {}^{2} }{2\pi r} \bigg)$

$\sf \large⇒ V = \frac{1}{2} [Sr - 2\pi r {}^{3} ]$

$\sf \large Now, \frac{dv}{dr} = \frac{1}{2} [S - 6\pi r {}^{2} ]$

$\sf \large⇒  \frac{d {}^{2}v }{dr {}^{2} } = \frac{1}{2} [0 - 12\pi r] = - 6\pi r$

$\sf \large For \frac{maximum}{minimum}$

$\sf \large \frac{dv}{dr} = 0$

$\sf \large⇒ S = 6\pi r {}^{2}$

From Equation (1)

$\sf \large⇒ 2\pi rh + 2\pi r {}^{2} = 6\pi r {}^{2}$

$\sf \large⇒ r = \frac{h}{2}$

$\sf \large[ \frac{d {}^{2}v }{dr {}^{2} } ]\sf\tt{\displaystyle \sf _{r = \frac{h}{2} }} \sf \large= - 6\pi \times \frac{h}{2} = - 3\pi h$

$\sf \large[ \frac{d {}^{2}v }{dr {}^{2} } ]\sf\tt{\displaystyle \sf _{r = \frac{h}{2} }} \sf \large< 0 \because - 3\pi h < 0$

$\sf \large⇒ volume \: of \: maximum \: at \: r = \frac{h}{2} .$

Answer:-

${ \boxed{ \sf \large \purple{Hence, h = 2r . }}}$

Hope you have satisfied. ⚘