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Question-213688

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Question Number 213688 by Abdullahrussell last updated on 13/Nov/24
Answered by mr W last updated on 13/Nov/24
at x_(min) =−2: −2+y=4(√(y+3)) y+3−4(√(y+3))−5=0 ((√(y+3))+1)((√(y+3))−5)=0 ⇒(√(y+3))=5 ⇒y=22 ⇒x+y=20 at y_(min) =−3: x−3=4(√(x+2)) x+2−4(√(x+2))−5=0 ((√(x+2))+1)((√(x+2))−5)=0 ⇒(√(x+2))=5 ⇒x=23 ⇒x+y=20 ⇒(x+y)_(min) =20 ✓
$${at}\:{x}_{{min}} =−\mathrm{2}: \\ $$$$−\mathrm{2}+{y}=\mathrm{4}\sqrt{{y}+\mathrm{3}} \\ $$$${y}+\mathrm{3}−\mathrm{4}\sqrt{{y}+\mathrm{3}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\sqrt{{y}+\mathrm{3}}+\mathrm{1}\right)\left(\sqrt{{y}+\mathrm{3}}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow\sqrt{{y}+\mathrm{3}}=\mathrm{5}\:\Rightarrow{y}=\mathrm{22}\:\Rightarrow{x}+{y}=\mathrm{20} \\ $$$${at}\:{y}_{{min}} =−\mathrm{3}: \\ $$$${x}−\mathrm{3}=\mathrm{4}\sqrt{{x}+\mathrm{2}} \\ $$$${x}+\mathrm{2}−\mathrm{4}\sqrt{{x}+\mathrm{2}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\sqrt{{x}+\mathrm{2}}+\mathrm{1}\right)\left(\sqrt{{x}+\mathrm{2}}−\mathrm{5}\right)=\mathrm{0} \\ $$$$\Rightarrow\sqrt{{x}+\mathrm{2}}=\mathrm{5}\:\Rightarrow{x}=\mathrm{23}\:\Rightarrow{x}+{y}=\mathrm{20} \\ $$$$\Rightarrow\left({x}+{y}\right)_{{min}} =\mathrm{20}\:\checkmark \\ $$
Answered by mr W last updated on 13/Nov/24
let p=x+2, q=y+3 p+q−5=4((√p)+(√q)) ((√p)+(√q))^2 −2(√(pq))−5=4((√p)+(√q)) ((√p)+(√q))^2 −4((√p)+(√q))−5=2(√(pq))≤p+q=4((√p)+(√q))+5 ((√p)+(√q))^2 −8((√p)+(√q))−10≤0 ⇒(√p)+(√q)≤4+(√(26)) ⇒x+y=4((√p)+(√q))≤4(4+(√(26))) ⇒(x+y)_(max) =4(4+(√(26))) ✓
$${let}\:{p}={x}+\mathrm{2},\:{q}={y}+\mathrm{3} \\ $$$${p}+{q}−\mathrm{5}=\mathrm{4}\left(\sqrt{{p}}+\sqrt{{q}}\right) \\ $$$$\left(\sqrt{{p}}+\sqrt{{q}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{{pq}}−\mathrm{5}=\mathrm{4}\left(\sqrt{{p}}+\sqrt{{q}}\right) \\ $$$$\left(\sqrt{{p}}+\sqrt{{q}}\right)^{\mathrm{2}} −\mathrm{4}\left(\sqrt{{p}}+\sqrt{{q}}\right)−\mathrm{5}=\mathrm{2}\sqrt{{pq}}\leqslant{p}+{q}=\mathrm{4}\left(\sqrt{{p}}+\sqrt{{q}}\right)+\mathrm{5} \\ $$$$\left(\sqrt{{p}}+\sqrt{{q}}\right)^{\mathrm{2}} −\mathrm{8}\left(\sqrt{{p}}+\sqrt{{q}}\right)−\mathrm{10}\leqslant\mathrm{0} \\ $$$$\Rightarrow\sqrt{{p}}+\sqrt{{q}}\leqslant\mathrm{4}+\sqrt{\mathrm{26}} \\ $$$$\Rightarrow{x}+{y}=\mathrm{4}\left(\sqrt{{p}}+\sqrt{{q}}\right)\leqslant\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{26}}\right) \\ $$$$\Rightarrow\left({x}+{y}\right)_{{max}} =\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{26}}\right)\:\checkmark \\ $$
Commented by mr W last updated on 14/Nov/24
alternatively let u=(√(x+2)), v=(√(y+3)) x+y=u^2 +v^2 −5=4(u+v) let u+v=k u^2 +(k−u)^2 −5=4k 2u^2 −2ku+k^2 −4k−5=0 Δ=k^2 −2(k^2 −4k−5)≥0 k^2 −8k−10≤0 ⇒k≤4+(√(26)) x+y=4k≤4(4+(√(26))) ⇒(x+y)_(max) =4(4+(√(26))) ✓
$${alternatively} \\ $$$${let}\:{u}=\sqrt{{x}+\mathrm{2}},\:{v}=\sqrt{{y}+\mathrm{3}} \\ $$$${x}+{y}={u}^{\mathrm{2}} +{v}^{\mathrm{2}} −\mathrm{5}=\mathrm{4}\left({u}+{v}\right) \\ $$$${let}\:{u}+{v}={k} \\ $$$${u}^{\mathrm{2}} +\left({k}−{u}\right)^{\mathrm{2}} −\mathrm{5}=\mathrm{4}{k} \\ $$$$\mathrm{2}{u}^{\mathrm{2}} −\mathrm{2}{ku}+{k}^{\mathrm{2}} −\mathrm{4}{k}−\mathrm{5}=\mathrm{0} \\ $$$$\Delta={k}^{\mathrm{2}} −\mathrm{2}\left({k}^{\mathrm{2}} −\mathrm{4}{k}−\mathrm{5}\right)\geqslant\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\mathrm{8}{k}−\mathrm{10}\leqslant\mathrm{0} \\ $$$$\Rightarrow{k}\leqslant\mathrm{4}+\sqrt{\mathrm{26}} \\ $$$${x}+{y}=\mathrm{4}{k}\leqslant\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{26}}\right) \\ $$$$\Rightarrow\left({x}+{y}\right)_{{max}} =\mathrm{4}\left(\mathrm{4}+\sqrt{\mathrm{26}}\right)\:\checkmark \\ $$

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