Question Number 122695 by liberty last updated on 19/Nov/20
$${Find}\:{maximum}\:{and}\:{minimum}\:{values} \\ $$$${of}\:{f}\left({x},{y}\right)\:=\:{xy}\:{on}\:{the}\:{circle}\:{g}\left({x},{y}\right)= \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:. \\ $$
Answered by john santu last updated on 19/Nov/20
$${By}\:{Lagrange}\:{multiplier}\: \\ $$$$\:\bigtriangledown{f}\left({p}\right)\:=\:\lambda\:\bigtriangledown{g}\left({p}\right)\:,{where}\:{P}\:{is}\:{any}\:{point} \\ $$$$\:\begin{pmatrix}{\frac{\partial{f}}{\partial{x}}}\\{\frac{\partial{f}}{\partial{y}}}\end{pmatrix}\:=\:\lambda\:\begin{pmatrix}{\frac{\partial{g}}{\partial{x}}}\\{\frac{\partial{g}}{\partial{y}}}\end{pmatrix} \\ $$$$\:\begin{pmatrix}{{y}}\\{{x}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}\lambda{x}}\\{\mathrm{2}\lambda{y}}\end{pmatrix}\:\rightarrow\begin{cases}{\lambda=\frac{{y}}{\mathrm{2}{x}}}\\{\lambda=\frac{{x}}{\mathrm{2}{y}}}\end{cases} \\ $$$$\:\Rightarrow\:\frac{{y}}{\mathrm{2}{x}}\:=\:\frac{{x}}{\mathrm{2}{y}}\:\Rightarrow{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \:\wedge{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\:{this}\:{gives}\:{four}\:{solutions}\: \\ $$$$\:\begin{cases}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:,\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}\\{\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:,\left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)}\end{cases} \\ $$$${Evaluating}\:{f}\:{at}\:{these}\:{four}\:{points} \\ $$$${we}\:{get}\:{max}\:{value}\:{is}\:\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{min}\:{value} \\ $$$${is}\:−\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
Answered by mr W last updated on 19/Nov/20
$${P}={xy} \\ $$$${y}=\frac{{P}}{{x}} \\ $$$${x}^{\mathrm{2}} +\frac{{P}^{\mathrm{2}} }{{x}^{\mathrm{2}} }−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}{x}+\frac{\mathrm{2}{PP}'}{{x}^{\mathrm{2}} }−\frac{\mathrm{2}{P}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$${P}'=\mathrm{0} \\ $$$${x}−\frac{{P}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$${x}−\frac{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }{{x}^{\mathrm{3}} }=\mathrm{0} \\ $$$${x}^{\mathrm{2}} ={y}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}},\:{y}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left({xy}\right)_{{max}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left({xy}\right)_{{min}} =−\frac{\mathrm{1}}{\mathrm{2}} \\ $$