Question Number 227247 by Spillover last updated on 10/Jan/26
Answered by Spillover last updated on 11/Jan/26
Answered by mr W last updated on 12/Jan/26
$$\frac{{r}}{{R}}=\frac{{h}−{y}}{{h}}\:\Rightarrow{y}={h}\left(\mathrm{1}−\frac{{r}}{{h}}\right) \\ $$$${V}=\pi{r}^{\mathrm{2}} {y}=\pi{hR}^{\mathrm{2}} \left(\mathrm{1}−\frac{{r}}{{R}}\right)\left(\frac{{r}}{{R}}\right)^{\mathrm{2}} =\pi{hR}^{\mathrm{2}} \left(\mathrm{1}−\xi\right)\xi^{\mathrm{2}} \\ $$$$\frac{{dV}}{{d}\xi}=\mathrm{0}\:\Rightarrow\mathrm{2}\xi−\mathrm{3}\xi^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\xi=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{y}={h}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{{h}}{\mathrm{3}}\:\checkmark \\ $$
Commented by Spillover last updated on 12/Jan/26
$${thanks} \\ $$