Question Number 137650 by bemath last updated on 05/Apr/21
Answered by bemath last updated on 05/Apr/21
$${By}\:{Langrange}\:{multiplier} \\ $$$${f}\left({x},{y},\lambda\right)=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\lambda\left(\mathrm{6}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$$\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{2}{x}+\lambda\left(\mathrm{12}{x}+\mathrm{2}{y}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}+\lambda\left(\mathrm{6}{x}+{y}\right)=\mathrm{0}\Rightarrow\lambda=−\frac{{x}}{\mathrm{6}{x}+{y}} \\ $$$$\frac{\partial{f}}{\partial{y}}=\mathrm{2}{y}+\lambda\left(\mathrm{2}{x}+\mathrm{12}{y}\right)=\mathrm{0} \\ $$$$\Rightarrow{y}+\lambda\left({x}+\mathrm{6}{y}\right)=\mathrm{0}\Rightarrow\lambda=−\frac{{y}}{{x}+\mathrm{6}{y}} \\ $$$$\Leftrightarrow\:\frac{{x}}{\mathrm{6}{x}+{y}}\:=\:\frac{{y}}{{x}+\mathrm{6}{y}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{6}{xy}={y}^{\mathrm{2}} +\mathrm{6}{xy} \\ $$$${it}\:{follows}\:{that}\:{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \:{or}\:\begin{cases}{{x}={y}}\\{{x}=−{y}}\end{cases} \\ $$$$\frac{\partial{f}}{\partial\lambda}=\:\mathrm{6}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow\mathrm{12}{x}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{9} \\ $$$${case}\left(\mathrm{1}\right){x}={y}\:\Rightarrow\mathrm{14}{x}^{\mathrm{2}} =\mathrm{9}\: \\ $$$${f}\left({x},{y}\right)\:=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:=\:\frac{\mathrm{9}}{\mathrm{7}} \\ $$$${case}\left(\mathrm{2}\right){x}=−{y}\:\Rightarrow\mathrm{10}{x}^{\mathrm{2}} =\mathrm{9} \\ $$$${f}\left({x},{y}\right)\:=\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$${Thus}\:{maximum}\:{value}\:{is}\:\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$ \\ $$
Commented by bemath last updated on 05/Apr/21
Answered by mr W last updated on 05/Apr/21
$${for}\:{a},{b}>\mathrm{0}\:{we}\:{have} \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\sqrt{{ab}} \\ $$$$ \\ $$$$\mathrm{9}=\mathrm{6}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{xy}\geqslant\mathrm{6}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{5}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\Rightarrow{maximum}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$ \\ $$$$\mathrm{9}=\mathrm{6}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{2}{xy}\leqslant\mathrm{6}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{7}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\frac{\mathrm{9}}{\mathrm{7}} \\ $$$$\Rightarrow{minimum}\:{of}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{7}} \\ $$
Commented by bemath last updated on 05/Apr/21
$${AM}−{GM} \\ $$