Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum?

If the sum of the surface areas of cube and a sphere is constant, what is the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum?

Answers (1)

Given: The combined surface area of a cube and sphere are constant

To find:  the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum

Explanation: Let ‘a’ be the side of the cube

Then surface area of the cube = 6a^2….(i)

Take ‘r’ as  the radius of the sphere

Then the surface area of the sphere = 4\pi r^2…(ii)

According to the question, the surface area of both the figures is added, thus adding the equation (i) and (ii), we get

\\ {6a\textsuperscript{2}+4 \pi r\textsuperscript{2} = k ($where k is the constant)}\\ { \Rightarrow 6a\textsuperscript{2} = k-4 \pi r\textsuperscript{2}}\\

\\\Rightarrow \mathrm{a}^{2}=\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}$ \\$\Rightarrow \mathrm{a}=\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}} \ldots$ (iii)
As the formula of volume of cube is
$\mathrm{v}_{c}=\mathrm{a}^{3}$
Plus the volume of a sphere is
\mathrm{V}_{\mathrm{s}}=\frac{4}{3} \pi \mathrm{r}^{3}$
Hence adding both the volumes will result into,
\mathrm{V}=\mathrm{a}^{3}+\frac{4}{3} \pi \mathrm{r}^{3}$

Then putting the values from equation (iii) in above equation,
$$ \begin{aligned} \mathrm{V} &=\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)^{3}+\frac{4}{3} \pi \mathrm{r}^{3} \\ \mathrm{~V} &=\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}+\frac{4}{3} \pi \mathrm{r}^{3} \end{aligned}
After finding the first derivative of the volume, we get
$$ \mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}+\frac{4}{3} \pi \mathrm{r}^{3}\right)}{\mathrm{dr}} $$
After taking out the constant terms along with using the sum rule of differentiating,
$$ \mathrm{V}^{\prime}=\frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}}\right)}{\mathrm{dr}}+\frac{4}{3} \pi \frac{\mathrm{d}\left(\mathrm{r}^{3}\right)}{\mathrm{dr}} $$

Using the power rule of differentiation,

\begin{aligned} \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{3}{2}-1} \cdot \frac{\mathrm{d}\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{\mathrm{dr}}+\frac{4}{3} \pi\left(3 \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[\frac{1}{6} \cdot \frac{\mathrm{d}\left(\mathrm{k}-4 \pi \mathrm{r}^{2}\right)}{\mathrm{dr}}\right]+\left(4 \pi \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[\frac{1}{6} \cdot(-4 \pi(2 \mathrm{r}))\right]+\left(4 \pi \mathrm{r}^{2}\right) \\ \mathrm{V}^{\prime} &=\frac{3}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\left[-\frac{4}{3} \pi r\right]+\left(4 \pi r^{2}\right) \\ \end{aligned}

\mathrm{V}^{\prime} =-2 \pi \mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}+\left(4 \pi \mathrm{r}^{2}\right)

\begin{aligned} &\mathrm{V}^{\prime}=-2 \pi \mathrm{r}\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right] \ldots . \text { (iv) }\\ &\text { Critical point can be calculated by equating the first derivative with } 0 \text { , }\\ &V^{\prime}=0\\ &\Rightarrow-2 \pi r\left[\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]=0\\ &\Rightarrow-2 \pi r=0 \text { or }\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]=0\\ &\Rightarrow \mathrm{r}=0 \text { or }\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}=2 \mathrm{r}\\ &\Rightarrow \mathrm{r}=0 \text { or } \frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}=4 \mathrm{r}^{2} \end{aligned}

\\ { \Rightarrow r = 0\ or\ (k-4 \pi r\textsuperscript{2}) = 24r\textsuperscript{2}}\\ { \Rightarrow r = 0\ or\ k = 4 \pi r\textsuperscript{2}+24r\textsuperscript{2}}\\ { \Rightarrow r = 0\ or\ (4 \pi +24)r\textsuperscript{2} = k}\\

\\\Rightarrow \mathrm{r}=0$ or $\mathrm{r}^{2}=\frac{\mathrm{k}}{(4 \pi+24)}$ \\$\Rightarrow \mathrm{r}=0$ or $\mathrm{r}=\pm \sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}$
Now we know, $r \neq 0$
Hence
$$ \mathrm{r}=\sqrt{\frac{\mathrm{k}}{(4 \pi+24)}} $$
To find the second derivative of this volume equation, we cam simply differentiate the equation (ii),
$$ \mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(-2 \pi \mathrm{r}\left[\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}-2 \mathrm{r}\right]\right)}{\mathrm{dx}} $$

$$ \mathrm{V}^{\prime \prime}=\frac{\mathrm{d}\left(-2 \pi \mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}+\left(4 \pi \mathrm{r}^{2}\right)\right)}{\mathrm{dx}} $$
After removing the constant terms, we apply the sum rule of differentiation,
$$ \mathrm{V}^{\prime \prime}=-2 \pi \frac{\mathrm{d}\left(\mathrm{r} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dx}} $$
Using the product rule of differentiation,
$$ \mathrm{V}^{\prime \prime}=-2 \pi\left[\mathrm{r} \frac{\mathrm{d}\left(\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right)}{\mathrm{dx}}+\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}} \cdot \frac{\mathrm{dr}}{\mathrm{dr}}\right]+4 \pi \frac{\mathrm{d}\left(\mathrm{r}^{2}\right)}{\mathrm{dx}} $$
Again, the power rule of differentiation is used,
$$ \mathrm{V}^{\prime \prime}=-2 \pi\left[\mathrm{r} \cdot \frac{1}{2} \cdot\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}-1} \frac{\mathrm{d}\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{\mathrm{dx}}+\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)^{\frac{1}{2}}\right]+4 \pi(2 \mathrm{r}) $$

Differentiating the equation, we get

V"=-2\pi\left [ r.\frac{1}{2}\left ( \frac{k-4\pi r^2}{6} \right )^\frac{-1}{2} \left ( \frac{1}{6}\left ( -4\pi(2r) \right ) \right )+\left ( \frac{k-4\pi r^2}{6} \right )^\frac{1}{2} \right ]+8\pi r

\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{-2 \pi r^{2}}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}+\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right]+8 \pi r

\\ \mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{-2 \pi r^{2}+3\left(\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}}{6}\right)}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \mathrm{\pi} \\ \left.\mathrm{~V}^{\prime \prime}=-2 \pi \left[ \frac{-2 \pi \mathrm{r}^{2}+\left(\frac{\mathrm{k}-4 \pi r^{2}}{2}\right)}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]\right]+8 \pi \mathrm{r}

\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{\frac{\mathrm{k}-4 \pi \mathrm{r}^{2}-4 \pi \mathrm{r}^{2}}{2}}{3\left(\left(\frac{\mathrm{k}-4 \pi r^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \pi \mathrm{r}

\mathrm{V}^{\prime \prime}=-2 \pi\left[\frac{\mathrm{k}-8 \pi \mathrm{r}^{2}}{6\left(\left(\frac{\mathrm{k}-4 \mathrm{\pi r}^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \mathrm{\pi r}

\mathrm{V}^{\prime \prime}=-\pi\left[\frac{\mathrm{k}-8 \mathrm{\pi r}^{2}}{6\left(\left(\frac{\mathrm{k}-4 \mathrm{\pi r}^{2}}{6}\right)^{\frac{1}{2}}\right)}\right]+8 \pi \mathrm{r}>0

r=\sqrt{\frac{k}{(4 \pi+24)}}$
Hence for
The substitutingr=\sqrt{\frac{k}{(4 \pi+24)}}, in equation (iii), we get
\Rightarrow \mathrm{a}=\left(\frac{\mathrm{k}-4 \pi\left(\sqrt{\frac{\mathrm{k}}{(4 \pi+24)}}\right)^{2}}{6}\right)^{\frac{1}{2}}$

\\ \Rightarrow a=\left(\frac{k-4 \pi\left(\frac{k}{4(\pi+6)}\right)}{6}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k-\left(\frac{k \pi}{(\pi+6)}\right)}{6}\right)^\frac{1}{2} \\ \Rightarrow a=\left(\frac{k(\pi+6)-k \pi}{6(\pi+6)}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k \pi+6 k-k \pi}{6(\pi+6)}\right)^{\frac{1}{2}} \\ \Rightarrow a=\left(\frac{k}{(\pi+6)}\right)^{\frac{1}{2}}

Now we will find the ratio of an edge of the cube to the diameter of the sphere, when the sums of their volumes are minimum, i.e.,

a:2r

\\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\left(\frac{\mathrm{k}}{(4 \pi+24)}\right)^{\frac{1}{2}}\right) \\ \\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\frac{1}{4^{\frac{1}{2}}}\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right) \\ \Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}: 2\left(\frac{1}{2}\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right)

\Rightarrow\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}:\left(\left(\frac{\mathrm{k}}{(\pi+6)}\right)^{\frac{1}{2}}\right)

Hence the required ratio is

a:2r = 1:1

 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question