Name: Rates of Change
Uploaded: 2019-07-19
Duration: 48 min 10 s
Description: In this video, let us know the average rate of change and instantaneous rate of change. Learn more about this concept in the video.

Question

Show that the height of a closed right circular cylinder of a given volume and least surface area is equal to its diameter.

Accepted Answer

Let $x$ be the radius of base, $h$ be the height and $S$ be the surface area of the closed right circular cylinder whose volume is $V$ which is given to be constant.

Then ${πr}^{2} h = V$

$∴ h = \frac{V}{{πr}^{2}} = \frac{A}{x^{2}} . . . . . . . . . (1)$

Where $A = \frac{V}{{πr}^{2}}$ , which is constant.

Now, $S = 2 πxh + 2 {πx}^{2}$

$S = 2 πx (\frac{A}{x^{2}}) + 2 {πx}^{2}$

$S = \frac{2 π A}{x} + 2 {πx}^{2} . . . . . . . . . (2)$

Differentiating $(2)$ w.r.t $x$ , we get

$∴ \frac{d S}{d x} = \frac{d}{d x} (\frac{2 πA}{x} + 2 {πx}^{2})$

$∴ \frac{d S}{d x} = 2 πA (- 1) x^{- 2} + 2 π \times 2 x$

$∴ \frac{d S}{d x} = \frac{- 2 πA}{x^{2}} + 4 πx . . . . . . . . . (3)$

Again differentiating $(3)$ w.r.t. $x$ , we get

$\frac{d^{2} S}{d x^{2}} = \frac{d}{d x} [\frac{- 2 πA}{x^{2}} + 4 πx]$

$\frac{d^{2} S}{d x^{2}} = - 2 πA (- 2) x^{- 3} + 4 π \times 1$

$\frac{d^{2} S}{d x^{2}} = \frac{4 πA}{x^{3}} + 4 π . . . . . . . . . . (4)$

Now, $\frac{d S}{d x} = 0$ gives $\frac{- 2 πA}{x^{2}} + 4 πx = 0$

$∴ - 2 π A + 4 {πx}^{3} = 0$

$∴ 4 {πx}^{3} = 2 π A$

$∴ x^{3} = \frac{A}{2}$

$∴ x = {(\frac{A}{2})}^{\frac{1}{3}}$

From $(4)$ ,

${(\frac{d^{2} S}{d x^{2}})}_{a t x = {(\frac{A}{2})}^{\frac{1}{3}}} = \frac{4 πA}{(\frac{A}{2})} + 4 π$

${(\frac{d^{2} S}{d x^{2}})}_{a t x = {(\frac{A}{2})}^{\frac{1}{3}}} = \frac{8 πA}{A} + 4 π = 12 π > 0$

$∴$ By the second derivative test, $S$ is minimum when $x = {(\frac{A}{2})}^{\frac{1}{3}}$ .

When $x = {(\frac{A}{2})}^{\frac{1}{3}}$ , from $(1)$ .

$h = \frac{A}{{(\frac{A}{2})}^{\frac{1}{3}}}$

$h = 2^{\frac{2}{3}} . A^{\frac{1}{3}}$

$h = 2^{(1 - \frac{1}{3})} \times A^{\frac{1}{3}}$

$h = \frac{2}{2^{\frac{1}{3}}} \times A^{\frac{1}{3}} = 2 \times {(\frac{A}{2})}^{\frac{1}{3}}$

$∴ h = 2 x$

Hence, the surface area is least when height of the closed right circular cylinder is equal to its diameter.

Show that the height of a closed right circular cylinder of a given volume and least surface area is equal to its diameter.

Important Questions on Applications of Derivatives

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