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Question-171727

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Question Number 171727 by cortano1 last updated on 20/Jun/22
Answered by mahdipoor last updated on 20/Jun/22
y≥((x^2 +y^2 )/2)⇒1≥x^2 +(y−1)^2 ⇒ −1≤x≤1 for x=x_0 , max(x_0 +y)=x_0 +max(y)= x_0 +1+(√(1−x_0 ^2 )) =f(x_0 ) (df/dx)=1−(x_0 /( (√(1−x_0 ^2 ))))=0 or ∄ ⇒x_0 =±1 , (√(1/2)) ⇒max(x+y)=f((√(1/2)))=2(√(1/2))+1
$${y}\geqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow\mathrm{1}\geqslant{x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{1} \\ $$$${for}\:{x}={x}_{\mathrm{0}} \:,\:{max}\left({x}_{\mathrm{0}} +{y}\right)={x}_{\mathrm{0}} +{max}\left({y}\right)= \\ $$$${x}_{\mathrm{0}} +\mathrm{1}+\sqrt{\mathrm{1}−{x}_{\mathrm{0}} ^{\mathrm{2}} }\:={f}\left({x}_{\mathrm{0}} \right) \\ $$$$\frac{{df}}{{dx}}=\mathrm{1}−\frac{{x}_{\mathrm{0}} }{\:\sqrt{\mathrm{1}−{x}_{\mathrm{0}} ^{\mathrm{2}} }}=\mathrm{0}\:{or}\:\nexists\:\Rightarrow{x}_{\mathrm{0}} =\pm\mathrm{1}\:,\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\Rightarrow{max}\left({x}+{y}\right)={f}\left(\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\right)=\mathrm{2}\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}+\mathrm{1} \\ $$
Answered by mr W last updated on 20/Jun/22
y≥((x^2 +y^2 )/2)=(((x+y)^2 )/2)−xy (x+y+1−y)y≥(((x+y)^2 )/2) let k=x+y (k+1)y−y^2 ≥(k^2 /2) y^2 −(k+1)y−(k^2 /2)≤0 Δ=(k+1)^2 −4×(k^2 /2)≥0 k^2 −2k−1≤0 (k−1)^2 ≤2 −(√2)+1≤k≤(√2)+1 (x+y)_(min) =k_(min) =−(√2)+1 (x+y)_(max) =k_(max) =(√2)+1
$${y}\geqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{2}}−{xy} \\ $$$$\left({x}+{y}+\mathrm{1}−{y}\right){y}\geqslant\frac{\left({x}+{y}\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$${let}\:{k}={x}+{y} \\ $$$$\left({k}+\mathrm{1}\right){y}−{y}^{\mathrm{2}} \geqslant\frac{{k}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\left({k}+\mathrm{1}\right){y}−\frac{{k}^{\mathrm{2}} }{\mathrm{2}}\leqslant\mathrm{0} \\ $$$$\Delta=\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}×\frac{{k}^{\mathrm{2}} }{\mathrm{2}}\geqslant\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{2}{k}−\mathrm{1}\leqslant\mathrm{0} \\ $$$$\left({k}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{2} \\ $$$$−\sqrt{\mathrm{2}}+\mathrm{1}\leqslant{k}\leqslant\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\left({x}+{y}\right)_{{min}} ={k}_{{min}} =−\sqrt{\mathrm{2}}+\mathrm{1} \\ $$$$\left({x}+{y}\right)_{{max}} ={k}_{{max}} =\sqrt{\mathrm{2}}+\mathrm{1} \\ $$
Answered by mr W last updated on 20/Jun/22
y≥((x^2 +y^2 )/2) x^2 +y^2 −2y≤0 x^2 +(y−1)^2 ≤1^2 → circle ⇒x=cos θ, y=1+sin θ x+y=1+sin θ+cos θ=1+(√2) sin (θ+(π/4)) (x+y)_(min) =1−(√2) (x+y)_(max) =1+(√2)
$${y}\geqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{y}\leqslant\mathrm{0} \\ $$$${x}^{\mathrm{2}} +\left({y}−\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{1}^{\mathrm{2}} \:\:\rightarrow\:{circle} \\ $$$$\Rightarrow{x}=\mathrm{cos}\:\theta,\:{y}=\mathrm{1}+\mathrm{sin}\:\theta \\ $$$${x}+{y}=\mathrm{1}+\mathrm{sin}\:\theta+\mathrm{cos}\:\theta=\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\left({x}+{y}\right)_{{min}} =\mathrm{1}−\sqrt{\mathrm{2}} \\ $$$$\left({x}+{y}\right)_{{max}} =\mathrm{1}+\sqrt{\mathrm{2}} \\ $$

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