Question

A cylindrical container is to be made from certain solid material with the following constraints:
It has a fixed inner volume of $Vm{{m}^{3}}$ , has a $2mm$ thick solid wall and is open at the top. The bottom of the container is a solid circular disc of thickness $2mm$ and is of radius equal to the outer radius of the container. If the volume of the material used to make the container is minimum when the inner radius of the container is$10mm$ , then the value of
$\dfrac{V}{250\pi }$ ​ Is

Answer 1

Hint: To solve the above question, we have to find the volume of the container. Let, if the inner radius of the container $r$ and height be $h$, then the volume of the container is $V=\pi {{r}^{2}}h$. After that we have to find the volume of material and then we have to find the minimum material required for the container. To find the minimum volume we will differentiate the volume w.r.t r and then we will equate the equation to 0.

Complete step-by-step answer:
If the inner radius of the container $r$ and height be $h$, then we have to get the volume of the container as $V=\pi {{r}^{2}}h$.
Now we have to find the volume of material that is $V=\pi {{\left( r+2 \right)}^{2}}\times h+\pi {{(r+2)}^{2}}\times 2-\pi {{r}^{2}}h$
Now we have to simplify it and then we will get,
$V=2\pi {{\left( r+2 \right)}^{2}}+\pi h\left( 4+4r \right)$
Now we are taking commons,
$V=2\pi {{\left( r+2 \right)}^{2}}+4\pi h\left( r+1 \right)$
Now,
$V=2\pi \left( {{\left( r+2 \right)}^{2}}+\dfrac{2\left( r+1 \right)V}{\pi {{r}^{3}}} \right)$
Now for minimum volume we have required $\dfrac{dv}{dr}=0$
Now we have to differentiate with respect to $r$
And we will get,
$\dfrac{dv}{dr}=2\pi \left( 2\left( r+2 \right)+\dfrac{2v}{\pi }\left( -\dfrac{1}{{{r}^{2}}}-\dfrac{2}{{{r}^{3}}} \right) \right)$
Now we have $\dfrac{dv}{dr}=0$
So,
$\Rightarrow 24+\dfrac{2V}{\pi }\left( \dfrac{-2-10}{{{10}^{3}}} \right)=0$
After simplifying we will get,
$\dfrac{24v}{{{10}^{3}}\pi }=24$
After omitting $24$ from both side we will get,
$v={{10}^{2}}\pi $
Now we will get,
$\dfrac{v}{250\pi }=4$
Hence the value of $\dfrac{v}{250\pi }$ is $4$ .


Note: Here student must take care of the formula Volume of container .Sometimes students makes
Mistakes on the derivative .So, they have to keep in mind some basic formulae $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ and also
Keep in mind that the constant term derivative is always zero, that is $\dfrac{d}{dx}\left( cons\tan t \right)=0$.