Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The sum of the surface areas of a rectangular parallelepiped with sides x, 2x and and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of

The sum of the surface areas of a rectangular parallelepiped with sides x, 2x and \frac{x}{3} and a sphere is given to be constant. Prove that the sum of their volumes is minimum, if x is equal to three times the radius of the sphere. Also find the minimum value of the sum of their volumes.

Answers (1)

Given: x, 2x and \frac{x}{3} are the sides of a rectangular parallelepiped. The sum of the surface areas of a sphere and a rectangular parallelepiped is given to be constant.

To prove: if x is equal to three times the radius of the sphere, the sum of their volumes is minimum and find the minimum value of the sum of their volumes

Surface area of rectangular parallelepiped:

\\ S_{\mathrm{rp}}=2\left(\mathrm{x}(2 \mathrm{x})+2 \mathrm{x}\left(\frac{\mathrm{x}}{3}\right)+\left(\frac{\mathrm{x}}{3}\right) \mathrm{x}\right) \\ \Rightarrow \mathrm{S}_{\mathrm{rp}}=2\left(2 \mathrm{x}^{2}+\frac{2 \mathrm{x}^{2}}{3}+\frac{\mathrm{x}^{2}}{3}\right)

Let radius of sphere be r cm, then surface area is S_s=4\pi r^2

Now sum of the surface areas is,

\mathrm{S}=\mathrm{S}_{\mathrm{rp}}+\mathrm{S}_{\mathrm{s}}=2\left(2 \mathrm{x}^{2}+\frac{2 \mathrm{x}^{2}}{3}+\frac{\mathrm{x}^{2}}{3}\right)+4 \pi \mathrm{r}^{2}

\Rightarrow \mathrm{S}=2\left(\frac{6 \mathrm{x}^{2}+2 \mathrm{x}^{2}+\mathrm{x}^{2}}{3}\right)+4 \mathrm{\pi}r^{2}

$\Rightarrow \mathrm{S}=6 \mathrm{x}^{2}+4 \mathrm{\pi}r^{2} \ldots$(i)
Now given that the sum of the surface areas is constant, so
\frac{\mathrm{dS}}{\mathrm{dr}}=0$
Now, differentiate (i) with respect to r and get
\frac{d S}{d r}=\frac{d\left(6 x^{2}+4 \pi r^{2}\right)}{d r}=0$
Apply differentiation rule of sum and get
\Rightarrow \frac{\mathrm{d}\left(6 \mathrm{x}^{2}\right)}{\mathrm{dr}}+\frac{\mathrm{d}\left(4 \pi \mathrm{r}^{2}\right)}{\mathrm{dr}}=0$
Take the constant terms out and get
\Rightarrow 6 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{dr}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dr}}=0$

Apply derivative and get
\\\Rightarrow 6 \times 2 \mathrm{x} \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dr}}+4 \pi \times 2 \mathrm{r} \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dr}}=0$ \\\\$\Rightarrow 12 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dr}}+8 \mathrm{\pi r}=0$\\ \\$\Rightarrow 12 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dr}}=-8 \pi \mathrm{r}$ \\\\$\Rightarrow \frac{d x}{d r}=-\frac{8 \pi r}{12 x}$ \\\\$\Rightarrow \frac{d x}{d r}=-\frac{2 \pi r}{3 x} \ldots \ldots . .$ (ii)
Let V denote the sum of volumes of both the shapes, so
\mathrm{V}=\mathrm{x} \times 2 \mathrm{x} \times \frac{\mathrm{x}}{3}+\frac{4}{3} \pi \mathrm{r}^{3}$

\Rightarrow \mathrm{V}=\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi \mathrm{r}^{3} \ldots \ldots$ (iii)
The first derivative of volume must be equal to 0 for minima or maxima
\frac{\mathrm{dV}}{\mathrm{dr}}=0$
Differentiate (iii) with respect to r and get
\frac{\mathrm{d} \mathrm{V}}{\mathrm{dr}}=\frac{\mathrm{d}\left(\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}}=0$
Apply differentiation rule of sum and get
\Rightarrow \frac{\mathrm{d}\left(\frac{2}{3} \mathrm{x}^{3}\right)}{\mathrm{dr}}+\frac{\mathrm{d}\left(\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}}=0$
Take constant terms out and get
\Rightarrow \frac{2}{3} \frac{\mathrm{d}\left(\mathrm{x}^{3}\right)}{\mathrm{dr}}+\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dr}}=0$
Apply derivative and get
\Rightarrow \frac{2}{3} \times 3 \mathrm{x}^{2} \times \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dr}}+\frac{4}{3} \pi \times 3 \mathrm{r}^{2} \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dr}}=0$

\Rightarrow 2 \mathrm{x}^{2} \frac{\mathrm{dx}}{\mathrm{dr}}+4 \mathrm{\pi}r^{2}=0$ $
Substitute value of \frac{d x}{d r}$ from (ii) and get
\\\Rightarrow 2 \mathrm{x}^{2}\left(-\frac{2 \pi r}{3 \mathrm{x}}\right)+4 \pi r^{2}=0$ \\$\Rightarrow-\frac{4 \pi r x}{3}+4 \pi r^{2}=0 \ldots . .(i v)$ \\$\Rightarrow 4 \pi r^{2}=\frac{4 \pi r x}{3}$ \\$\Rightarrow \mathrm{r}=\frac{\mathrm{x}}{3}$ $\Rightarrow x=3 r$
i.e., the radius of the sphere is 1/3 of x.

Hence proved

Now let’s find the second derivative value at x=3r.

Now, apply derivative with respect to r to (iv) and get

\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=\frac{\mathrm{d}\left(-\frac{4 \pi \mathrm{r} \mathrm{x}}{3}+4 \mathrm{\pi r}^{2}\right)}{\mathrm{dr}}$
Apply differentiation rule of sum and get
\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=\frac{\mathrm{d}\left(-\frac{4 \pi \mathrm{rx}}{3}\right)}{\mathrm{dr}}+\frac{\mathrm{d}\left(4 \mathrm{\pi r}^{2}\right)}{\mathrm{dr}}$
Take constant terms out and get
\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3} \frac{\mathrm{d}(\mathrm{rx})}{\mathrm{dr}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dr}}$
The first part is applied the differentiation rule of product, so

\\\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(\mathrm{r} \frac{\mathrm{dx}}{\mathrm{dr}}+\mathrm{x} \frac{\mathrm{dr}}{\mathrm{dr}}\right)+4 \pi \times 2 \mathrm{r} \frac{\mathrm{d}(\mathrm{r})}{\mathrm{dr}}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(\mathrm{r} \frac{\mathrm{d} \mathrm{x}}{\mathrm{dr}}+\mathrm{x}\right)+4 \pi \times 2 \mathrm{r}$

 Substitute value of \frac{dr}{dx} from (ii) and get
\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(\mathrm{r}\left(-\frac{2 \pi \mathrm{r}}{3 \mathrm{x}}\right)+\mathrm{x}\right)+4 \pi \times 2 \mathrm{r}$ \\$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}=-\frac{4 \pi}{3}\left(-\frac{2 \pi \mathrm{r}^{2}}{3 \mathrm{x}}+\mathrm{x}\right)+8 \mathrm{\pi r}$
Substitutex=3r and get
\\\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=-\frac{4 \pi}{3}\left(-\frac{2 \pi \mathrm{r}^{2}}{3(3 \mathrm{r})}+3 \mathrm{r}\right)+8 \mathrm{\pi r}$ \\$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=\frac{4 \pi}{3}\left(\frac{2\pi \mathrm{r}}{9}-3 \mathrm{r}\right)+8 \mathrm{\pi r}$ \\$\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=\frac{4 \pi}{3}\left(\frac{2\pi \mathrm{r}}{9}-3 \mathrm{r}+6 \mathrm{r}\right)$

\Rightarrow\left(\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{dr}^{2}}\right)_{\mathrm{x}=3 \mathrm{r}}=\frac{4 \pi}{3}\left(\frac{\pi \mathrm{r}}{3}+3 \mathrm{r}\right)$
It is positive;so  V is minimum when \mathrm{x}=3 \mathrm{r}$ or $\mathrm{r}=\frac{\mathrm{x}}{3}, and the minimum value of Volume
can be obtained by substituting r=\frac{x}{3}$ in equation (iii), we get
\\\mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}$ \\$\Rightarrow \mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2}{3}(\mathrm{x})^{3}+\frac{4}{3} \pi\left(\frac{\mathrm{x}}{3}\right)^{3}$ \\$\Rightarrow \mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2}{3} \mathrm{x}^{3}+\frac{4}{3} \pi\left(\frac{\mathrm{x}^{3}}{27}\right)$ \\$\Rightarrow \mathrm{V}_{\mathrm{r}=\frac{\mathrm{x}}{3}}=\frac{2 \mathrm{x}^{3}}{3}\left[1+\frac{2 \pi}{27}\right]$

Therefore, it Is the minimum value of the sum of their volumes.

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question