Question Number 212500 by ajfour last updated on 15/Oct/24
$${Can}\:{we}\:{exactly}\:{find}\:{r}_{{max}} \left({a}\right). \\ $$$${r}^{\mathrm{2}} +\frac{\mathrm{2}{rt}}{{a}}\left({t}+\mathrm{2}\sqrt{{ar}}\right)={t}^{\mathrm{2}} \:\:\:\:\forall\:{t}\:{is}\:{parameter} \\ $$
Answered by mr W last updated on 15/Oct/24
$$\left(\frac{\mathrm{2}{r}}{{a}}−\mathrm{1}\right){t}^{\mathrm{2}} +\mathrm{4}{r}\sqrt{\frac{{r}}{{a}}}{t}+{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\frac{\mathrm{4}{r}^{\mathrm{3}} }{{a}}−{r}^{\mathrm{2}} \left(\frac{\mathrm{2}{r}}{{a}}−\mathrm{1}\right)\geqslant\mathrm{0} \\ $$$$\left(\frac{\mathrm{2}{r}}{{a}}+\mathrm{1}\right){r}^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\frac{\mathrm{2}{r}}{{a}}+\mathrm{1}\geqslant\mathrm{0}\: \\ $$$${this}\:{is}\:{always}\:{true},\:{since}\:{ar}\geqslant\mathrm{0}. \\ $$$${if}\:{a}>\mathrm{0}:\:{r}\geqslant\mathrm{0} \\ $$$${if}\:{a}<\mathrm{0}:\:{r}\leqslant\mathrm{0} \\ $$
Commented by ajfour last updated on 15/Oct/24
$${True}!\:{not}\:{your}\:{fault}. \\ $$