Question

please answer fast .....​

Answer 1

Let $\sf{d=2r}$ be the diameter of the semi - circular ends where $\sf{r}$ is the radius, and $\sf{h}$ be the height of the half - cylinder.

Then by question, we have,

$\displaystyle\longrightarrow\sf{\dfrac{h}{d}=\dfrac{\pi}{\pi+k}\quad\quad\dots(1)}$

Here the volume of the metal is unchanged even after casting, and is given by,

$\displaystyle\longrightarrow\sf{V=\dfrac{1}{2}\,\pi r^2h}$

$\displaystyle\longrightarrow\sf{V=\dfrac{1}{2}\,\pi\left(\dfrac{d}{2}\right)^2h}$

$\displaystyle\longrightarrow\sf{V=\dfrac{1}{2}\,\pi\cdot\dfrac{d^2}{4}\cdot h}$

$\displaystyle\longrightarrow\sf{V=\dfrac{1}{8}\,\pi d^2h}$

Then we get the following proportionality,

$\displaystyle\longrightarrow\sf{h\propto\dfrac{1}{d^2}\quad\quad\dots(2)}$

But we consider the equality here actually, to avoid complexities, if any.

The total surface area of the half cylinder is equal to the sum of areas of the two semi - circular ends, the curved surface and the plane surface. Thus,

$\displaystyle\longrightarrow\sf{A=2\cdot\dfrac{\pi r^2}{2}+\dfrac{2\pi r}{2}\cdot h+2rh}$

$\displaystyle\longrightarrow\sf{A=\pi r^2+\pi rh+2rh}$

$\displaystyle\longrightarrow\sf{A=\pi\left(\dfrac{d}{2}\right)^2+\pi\left(\dfrac{d}{2}\right)h+2\left(\dfrac{d}{2}\right)h}$

$\displaystyle\longrightarrow\sf{A=\dfrac{\pi d^2}{4}+\dfrac{\pi dh}{2}+dh}$

Using (2) we are going to convert $\sf{h}$ in terms of $\sf{d}$ so that $\sf{A}$ has to be written in terms of $\sf{d}$ only. Similarly $\sf{d}$ can be converted to $\sf{h}$ too. Thus,

$\displaystyle\longrightarrow\sf{A=\dfrac{\pi d^2}{4}+\dfrac{\pi d}{2d^2}+\dfrac{d}{d^2}}$

$\displaystyle\longrightarrow\sf{A=\dfrac{\pi d^2}{4}+\dfrac{\pi}{2d}+\dfrac{1}{d}}$

For minimum $\sf{A,}$ its derivative with respect to $\sf{d}$ should be taken as zero.

$\displaystyle\longrightarrow\sf{\dfrac{dA}{dd}=0}$

$\displaystyle\longrightarrow\sf{\dfrac{d}{dd}\,\left[\dfrac{\pi d^2}{4}+\dfrac{\pi}{2d}+\dfrac{1}{d}\right]=0}$

Since $\sf{\dfrac{d}{dx}\left[x^2\right]=2x}$ and $\sf{\dfrac{d}{dx}\left[\dfrac{1}{x}\right]=-\dfrac{1}{x^2},}$

$\displaystyle\longrightarrow\sf{\dfrac{\pi d}{2}-\dfrac{\pi}{2d^2}-\dfrac{1}{d^2}=0}$

$\displaystyle\longrightarrow\sf{\dfrac{\pi d^3-\pi-2}{2d^2}=0}$

$\displaystyle\longrightarrow\sf{\pi d^3-\pi-2=0}$

$\displaystyle\longrightarrow\sf{d^3=\dfrac{\pi+2}{\pi}}$

$\displaystyle\longrightarrow\sf{d\cdot d^2=\dfrac{\pi+2}{\pi}}$

From (2), [equality is taken]

$\displaystyle\longrightarrow\sf{\dfrac{d}{h}=\dfrac{\pi+2}{\pi}}$

$\displaystyle\longrightarrow\sf{\dfrac{h}{d}=\dfrac{\pi}{\pi+2}\quad\quad\dots(3)}$

By comparing (1) and (3), we get,

$\displaystyle\longrightarrow\sf{\underline{\underline{k=2}}}$