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A given quantity of metal is to be cast into a half cylinder (i,e, with a rectangular base and semicircular ends). If the total surface area is to be minimum, then the ratio of the height of the half cylinder to the diameter of the semicircular ends is
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The correct answer is:
$\pi:(\pi+2)$
Volume of half cylinder with height ${ }^{\prime} 1^{\prime}$ and radius ' $\mathrm{r}$ '
$=\frac{1}{2} \pi r^{2} l$
Surface area of half cylinder.
$S=l \times 2 r+2\left(\frac{1}{2} \pi r^{2}\right)+2\left(\frac{1}{2} \pi r l\right)$
$S=2 r l+\pi r^{2}+\pi r l$
$S=\frac{4 V}{\pi r}+\frac{\pi \cdot 2 V}{\pi r}+\pi r^{2} \quad($ from $(1))$
$\frac{d s}{d r} \Rightarrow \frac{-4 V}{\pi r^{2}}-\frac{2 V}{r^{2}}+2 \pi r=0$
$\Rightarrow 2 \pi r=\frac{4 V}{\pi r^{2}}+\frac{2 V}{r^{2}}$
$\Rightarrow 2 \pi r=\frac{4 . \pi r^{2} l}{2 \pi r^{2}}+\frac{2}{r^{2}} \cdot \frac{1}{2} \pi r^{2} l$
$\Rightarrow \frac{l}{2 r}=\frac{\pi}{\pi+2}$
$=\frac{1}{2} \pi r^{2} l$
Surface area of half cylinder.
$S=l \times 2 r+2\left(\frac{1}{2} \pi r^{2}\right)+2\left(\frac{1}{2} \pi r l\right)$
$S=2 r l+\pi r^{2}+\pi r l$
$S=\frac{4 V}{\pi r}+\frac{\pi \cdot 2 V}{\pi r}+\pi r^{2} \quad($ from $(1))$
$\frac{d s}{d r} \Rightarrow \frac{-4 V}{\pi r^{2}}-\frac{2 V}{r^{2}}+2 \pi r=0$
$\Rightarrow 2 \pi r=\frac{4 V}{\pi r^{2}}+\frac{2 V}{r^{2}}$
$\Rightarrow 2 \pi r=\frac{4 . \pi r^{2} l}{2 \pi r^{2}}+\frac{2}{r^{2}} \cdot \frac{1}{2} \pi r^{2} l$
$\Rightarrow \frac{l}{2 r}=\frac{\pi}{\pi+2}$
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