Question Number 219890 by SdC355 last updated on 03/May/25
$$\mathrm{Find}\:\mathrm{Maxima}\: \\ $$$${x}+{y}\:\mathrm{where}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:\left(\mathrm{use}\:\mathrm{Lagrange}\:\mathrm{Method}\right) \\ $$
Answered by MrGaster last updated on 03/May/25
$$\bigtriangledown\left({x}+{y}\right)=\lambda\bigtriangledown\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\left(\mathrm{1},\mathrm{1}\right)=\lambda\left(\mathrm{2}{x},\mathrm{2}{y}\right) \\ $$$$\Rightarrow\mathrm{1}=\mathrm{2}\lambda{x}\:\wedge\:\mathrm{1}=\mathrm{2}\lambda{y} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{r}^{\mathrm{2}} =\Rightarrow\mathrm{2}{x}^{\mathrm{2}} ={r}^{\mathrm{2}} \Rightarrow{x}=\pm\frac{{r}}{\:\sqrt{\mathrm{2}}} \\ $$$$\Rightarrow{x}+{y}=\pm\frac{{r}}{\:\sqrt{\mathrm{2}}}\pm\frac{{r}}{\:\sqrt{\mathrm{2}}}=\pm\sqrt{\mathrm{2}}{r} \\ $$$$\Rightarrow\mathrm{max}\left({x}+{y}\right)=\sqrt{\mathrm{2}}{r} \\ $$