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Question-197636




Question Number 197636 by universe last updated on 25/Sep/23
Answered by mr W last updated on 25/Sep/23
s=x+y  (x+3)^2 +y^2 =(8(√2))^2   (x+3)^2 +(s−x)^2 =128  2(x+3)+2(s−x)(−1)=0  ⇒s=2x+3  (x+3)^2 +(2x+3−x)^2 =128  ⇒x+3=8   ⇒x=5   ⇒s_(max) =2×5+3=13 ✓
$${s}={x}+{y} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +{y}^{\mathrm{2}} =\left(\mathrm{8}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left({s}−{x}\right)^{\mathrm{2}} =\mathrm{128} \\ $$$$\mathrm{2}\left({x}+\mathrm{3}\right)+\mathrm{2}\left({s}−{x}\right)\left(−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{s}=\mathrm{2}{x}+\mathrm{3} \\ $$$$\left({x}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{2}{x}+\mathrm{3}−{x}\right)^{\mathrm{2}} =\mathrm{128} \\ $$$$\Rightarrow{x}+\mathrm{3}=\mathrm{8}\: \\ $$$$\Rightarrow{x}=\mathrm{5}\: \\ $$$$\Rightarrow{s}_{{max}} =\mathrm{2}×\mathrm{5}+\mathrm{3}=\mathrm{13}\:\checkmark \\ $$
Commented by universe last updated on 25/Sep/23
thanks sir
$${thanks}\:{sir} \\ $$

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