Question Number 174538 by infinityaction last updated on 03/Aug/22
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and} \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:``\mathrm{x}''\:\mathrm{if} \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{6},\:\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{18}, \\ $$$$\mathrm{for}\:\mathrm{real}\:\mathrm{values}\:\mathrm{of}\:``\mathrm{x},\mathrm{y}\:\mathrm{and}\:\mathrm{z}'' \\ $$
Commented by JDamian last updated on 03/Aug/22
isn't the intersection of a 3D straight line with a centered sphere?
Commented by mr W last updated on 03/Aug/22
$$\mathrm{0}\leqslant{x},{y},{z}\leqslant\mathrm{4} \\ $$
Commented by infinityaction last updated on 03/Aug/22
$${yes}\:{sir}\: \\ $$$${share}\:{your}\:{solution} \\ $$
Answered by mr W last updated on 03/Aug/22
$${z}=\mathrm{6}−{x}−{y} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\left(\mathrm{6}−{x}−{y}\right)^{\mathrm{2}} =\mathrm{18} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{y}^{\mathrm{2}} −\mathrm{12}{x}−\mathrm{12}{y}+\mathrm{2}{xy}+\mathrm{18}=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{6}−{x}\right){y}+\left(\mathrm{3}−{x}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\mathrm{6}−{x}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{3}−{x}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$${x}\left(\mathrm{4}−{x}\right)\geqslant\mathrm{0} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}\leqslant\mathrm{4} \\ $$$${similarly} \\ $$$$\Rightarrow\mathrm{0}\leqslant{y}\leqslant\mathrm{4} \\ $$$$\Rightarrow\mathrm{0}\leqslant{z}\leqslant\mathrm{4} \\ $$
Commented by infinityaction last updated on 03/Aug/22
$${thanks}\:{sir} \\ $$