Question Number 118701 by bemath last updated on 19/Oct/20
$${What}\:{is}\:{maximum}\:{value} \\ $$$$\:{x}^{\mathrm{3}} {y}^{\mathrm{3}} \:+\mathrm{3}{xy}\:{when}\:{x}+{y}\:=\:\mathrm{8}? \\ $$
Commented by PRITHWISH SEN 2 last updated on 19/Oct/20
$$\mathrm{4144} \\ $$
Commented by bemath last updated on 19/Oct/20
$${x}={y}=\mathrm{4}\: \\ $$$$\left(\mathrm{16}\right)^{\mathrm{3}} +\mathrm{3}×\mathrm{4}×\mathrm{4}\:=\:\mathrm{4144} \\ $$$$ \\ $$
Answered by mr W last updated on 19/Oct/20
$$\left({xy}\right)^{\mathrm{3}} +\mathrm{3}\left({xy}\right) \\ $$$$=\left(\sqrt{{xy}}\right)^{\mathrm{6}} +\mathrm{3}\left(\sqrt{{xy}}\right)^{\mathrm{2}} \\ $$$$\leqslant\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{6}} +\mathrm{3}\left(\frac{{x}+{y}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{6}} +\mathrm{3}×\mathrm{4}^{\mathrm{2}} =\mathrm{4144} \\ $$