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Question Number 59620 by ajfour last updated on 12/May/19

Commented by ajfour last updated on 12/May/19

$$\mathrm{Find}\mathrm{dimensions}\mathrm{of}\mathrm{largest}\mathrm{volume}$$$$\mathrm{cylinder}\mathrm{inscribed}\mathrm{within}\mathrm{a}\mathrm{hemi}-$$$$\mathrm{sphere}\mathrm{of}\mathrm{radius}\mathrm{R}\left(\mathrm{in}\mathrm{shown}\mathrm{orientation}\right).$$

Commented by MJS last updated on 12/May/19

$$\mathrm{more}\mathrm{difficult}:\mathrm{find}\mathrm{the}\mathrm{cylinder}\mathrm{with}\mathrm{the}$$$$\mathrm{largest}\mathrm{surface}$$$$\Rightarrow$$$$l=\frac{\sqrt{34(17-\sqrt{17})}}{17}R$$$$r=\frac{\sqrt{34(17+\sqrt{17})}}{68}R$$$$S=\frac{\pi }{4}(1+\sqrt{17})R^2$$

Answered by MJS last updated on 12/May/19

$$R^2-\frac{l^2}{4}=4r^2\Rightarrow r=\frac{1}{4}\sqrt{4R^2-l^2}$$$$V=\pi {lr}^2=\frac{\pi }{4}l\sqrt{4R^2-l^2}$$$$\frac{dV}{dl}=\frac{\pi }{2}\times \frac{2R^2-l^2}{\sqrt{4R^2-l^2}}=0\Rightarrow l=\sqrt{2}R\Rightarrow r=\frac{\sqrt{2}}{4}R\Rightarrow$$$$\Rightarrow V_{max}=\frac{\pi \sqrt{2}}{8}{R}^3$$

Commented by ajfour last updated on 13/May/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Sir}!$$

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