Question

A given quantity of metal is to be cast into a solid half circular cylinder with a rectangular base and semi-circular ends. Show that in order that total surface area is minimum, the ratio of length of cylinder to the diameter of semi-circular ends is π ∶ π + 2.

Answer 1

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Answer 2

Ratio of the length of the cylinder to the diameter of its semi circular:

The cylinder with semi-circle is given in the image below.

Let's consider,

l = Length of the circular cylinder

r = Radius of semicircular in the ends of the circular cylinder

Volume of the cylinder is:

$\bold{V=\frac{1}{2} \pi r^{2} h}$

Curved area of half cylinder is:

$\bold{C=\frac{1}{2} (2 \pi r h)}$

Area of rectangular base is:

$\bold{A=2 rh}$

Area of semicircular ends is:

$\bold{SM=\frac{1}{2} \pi r^{2}}$

Total surface area of the half cylinder is:

$\bold{S=(2\times SM)+C+A}$

On substituting the formula, we get,

$\bold{S=(2 \times \frac{1}{2} \pi r^{2})+(\frac{1}{2} (2 \pi r h))+(2 r h)}$

$\bold{S=\pi r^{2}+(\pi+2) r h}$

$\bold{S=\pi r^{2}+(\pi+2) r \times \frac{2 V}{\pi r^{2}}}$

$\bold{S=\pi r^{2}+\frac{2 V(\pi+2)}{\pi r}}$

On differentiating above equation with respect to 'r', we get,

$\bold{\frac{d S}{d r}=2 \pi r+\frac{2 V(\pi+2)}{\pi}\left(-\frac{1}{r^{2}}\right)}$

Again differentiating with respect to 'r', we get,

$\bold{\frac{d^{2} S}{d r^{2}}=2 \pi+\frac{2 V(\pi+2)}{\pi} \frac{2}{r^{3}}}$

Now,

$\bold{\frac{d S}{d r}=0}$

$\bold{2 \pi r-\frac{2(\pi+2) V}{\pi r^{2}}=0}$

$\bold{r^{3}=\frac{(\pi+2) V}{\pi^{2}}}$

On substituting $\bold{r^3}$ in equation differentiated for second, we get,

$\bold{\frac{d^{2} S}{d r^{2}}=2 \pi+\frac{4}{\pi} \times \pi^{2}=6 \pi>0}$

The total surface area is minimum when r is:

$\bold{r^{3}=\frac{(\pi+2) V}{\pi^{2}}}$

On substituting volume, we get,

$\bold{r^{3}=\frac{\pi+2}{\pi^{2}} \times \frac{1}{2} \pi r^{2} h}$

$\bold{r=\frac{\pi+2}{2 \pi} \times h}$

$\bold{\frac{h}{2 r}=\frac{\pi}{\pi+2}}$

h = length of cylinder

2r = diameter of its circular base

Hence proved.