Question Number 219570 by Rojarani last updated on 29/Apr/25
$$\:\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} =\mathrm{36}\left({x}+{y}\right),\: \\ $$$$\:\:{x},{y}\in{R},\:{find}\:{maximum}\:\left({x}+{y}\right) \\ $$$$\: \\ $$
Answered by SdC355 last updated on 29/Apr/25
$$\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}{y}^{\mathrm{2}} =\mathrm{36}\left({x}+{y}\right)\:\overset{\mathrm{Equal}} {\rightleftarrows}\:\frac{\mathrm{16}}{\mathrm{153}}\left({x}−\frac{\mathrm{9}}{\mathrm{4}}\right)^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{17}}\left({y}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${x}=\frac{\mathrm{9}}{\mathrm{4}}\:,\:{y}=\mathrm{2}\: \\ $$$${x}+{y}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$
Answered by mr W last updated on 29/Apr/25
$${say}\:{x}+{y}={s}\: \\ $$$$\mathrm{8}{x}^{\mathrm{2}} +\mathrm{9}\left({s}−{x}\right)^{\mathrm{2}} −\mathrm{36}{s}=\mathrm{0} \\ $$$$\mathrm{17}{x}^{\mathrm{2}} −\mathrm{18}{sx}+\mathrm{9}{s}^{\mathrm{2}} −\mathrm{36}{s}=\mathrm{0} \\ $$$$\Delta=\mathrm{18}^{\mathrm{2}} {s}^{\mathrm{2}} −\mathrm{4}×\mathrm{17}\left(\mathrm{9}{s}^{\mathrm{2}} −\mathrm{36}{s}\right)\geqslant\mathrm{0} \\ $$$$\left(\mathrm{17}−\mathrm{2}{s}\right){s}\geqslant\mathrm{0} \\ $$$$\Rightarrow{s}\geqslant\mathrm{0},\:\mathrm{17}−\mathrm{2}{s}\geqslant\mathrm{0}\:\Rightarrow{s}\leqslant\frac{\mathrm{17}}{\mathrm{2}}\:\Rightarrow\mathrm{0}\leqslant{x}+{y}=\frac{\mathrm{17}}{\mathrm{2}} \\ $$$${i}.{e}.\left({x}+{y}\right)_{{max}} =\frac{\mathrm{17}}{\mathrm{2}},\:\left({x}+{y}\right)_{{min}} =\mathrm{0} \\ $$
Commented by mr W last updated on 29/Apr/25
