Question Number 137743 by bemath last updated on 06/Apr/21
$$ \\ $$ find the max and min of f(x,y)=81x^2+y^2, subject 4x^2+y^2=9
Answered by EDWIN88 last updated on 06/Apr/21
$${by}\:{Lagrange}\:{multiplier} \\ $$$${f}\left({x},{y},\lambda\right)=\mathrm{81}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\lambda\left(\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$$\frac{{df}}{{dx}}\:=\:\mathrm{162}{x}+\mathrm{8}\lambda{x}\:=\:\mathrm{0}\:;\:\lambda=−\frac{\mathrm{162}}{\mathrm{8}}=−\frac{\mathrm{81}}{\mathrm{4}} \\ $$$$\frac{{df}}{{dy}}=\mathrm{2}{y}+\mathrm{2}\lambda{y}=\mathrm{0};\:\lambda=−\mathrm{1} \\ $$$${for}\:\lambda=−\frac{\mathrm{81}}{\mathrm{4}}\:\Rightarrow{f}\left({x},{y}\right)=\mathrm{81}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\frac{\mathrm{81}}{\mathrm{4}}\left(\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$${f}\left({x},{y}\right)=−\frac{\mathrm{77}}{\mathrm{4}}{y}^{\mathrm{2}} +\frac{\mathrm{729}}{\mathrm{4}}\:\Rightarrow{max}\:=\:\frac{\mathrm{729}}{\mathrm{4}}=\mathrm{182}.\mathrm{25} \\ $$$$ \\ $$$${for}\:\lambda=−\mathrm{1}\Rightarrow{f}\left({x},{y}\right)=\mathrm{81}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{1}.\left(\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$${f}\left({x},{y}\right)=\mathrm{77}{x}^{\mathrm{2}} +\mathrm{9}\:\Rightarrow{min}\:=\:\mathrm{9} \\ $$
Answered by mr W last updated on 06/Apr/21
$$\mathrm{4}{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{9} \\ $$$$\Rightarrow\mathrm{0}\leqslant{x}^{\mathrm{2}} \leqslant\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${f}\left({x},{y}\right)=\mathrm{81}{x}^{\mathrm{2}} +\mathrm{9}−\mathrm{4}{x}^{\mathrm{2}} =\mathrm{77}{x}^{\mathrm{2}} +\mathrm{9} \\ $$$$\geqslant\mathrm{77}×\mathrm{0}+\mathrm{9}=\mathrm{9}\:\Rightarrow{minimum} \\ $$$$\leqslant\mathrm{77}×\frac{\mathrm{9}}{\mathrm{4}}+\mathrm{9}=\mathrm{182}.\mathrm{25}\:\Rightarrow{maximum} \\ $$
Commented by mr W last updated on 06/Apr/21
