Question Number 115922 by mathdave last updated on 29/Sep/20
$${find}\:{the}\:{stationary}\:{points}\:{of}\:{the} \\ $$$${function}\:{U}={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{subjects}\:{to} \\ $$$${the}\:{constraint}\: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$
Answered by bemath last updated on 29/Sep/20
$${f}\left({x},{y},\lambda\right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\lambda\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}{y}+\mathrm{1}\right) \\ $$$$\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{2}{x}+\lambda\left(\mathrm{2}{x}+\mathrm{2}\right)=\mathrm{0}\rightarrow\lambda=\frac{−\mathrm{2}{x}}{\mathrm{2}{x}+\mathrm{2}} \\ $$$$\frac{\partial{f}}{\partial{y}}\:=\:\mathrm{2}{y}+\lambda\left(\mathrm{2}{y}−\mathrm{2}\right)=\mathrm{0}\rightarrow\lambda=\frac{\mathrm{2}{y}}{\mathrm{2}−\mathrm{2}{y}} \\ $$$$\frac{\partial{f}}{\partial\lambda}=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}{y}+\mathrm{1}=\mathrm{0} \\ $$$$\Leftrightarrow\:\frac{{x}}{\mathrm{2}{x}+\mathrm{2}}\:=\:\frac{{y}}{\mathrm{2}{y}−\mathrm{2}}\:\Rightarrow\mathrm{2}{xy}−\mathrm{2}{x}=\mathrm{2}{xy}+\mathrm{2}{y} \\ $$$${y}=−{x}\:\Rightarrow{x}^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}=\mathrm{0}\:;\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}\:\Rightarrow{x}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1} \\ $$$${and}\:{y}\:=\:\mp\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{1} \\ $$$$ \\ $$