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Question-205338

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Question Number 205338 by cortano12 last updated on 17/Mar/24
Answered by mr W last updated on 17/Mar/24
x≥−2, y≥−3 x+2−4(√(x+2))+4+y+3−4(√(y+3))+4=13 ((√(x+2))−2)^2 +((√(y+3))−2)^2 =((√(13)))^2 (√(x+2))−2=(√(13)) cos θ (√(y+3))−2=(√(13)) sin θ ⇒x=(2+(√(13)) cos θ)^2 −2=2+4(√(13)) cos θ+13 cos^2 θ ⇒y=(2+(√(13)) sin θ)^2 −3=1+4(√(13)) sin θ+13 sin^2 θ x+y=16+4(√(13))(cos θ+sin θ) x+y=16+4(√(26)) sin (θ+(π/4)) ⇒(x+y)_(max) =16+4(√(26)) ✓ at x=−2: ((√(y+3))−2)^2 =13−(−2)^2 =9 (√(y+3))−2=3 y=22 ⇒x+y=−2+22=20 at y=−3: ((√(x+2))−2)^2 =13−(−2)^2 =9 (√(x+2))−2=3 x=23 ⇒x+y=23−3=20 ⇒(x+y)_(min) =20 ✓
$${x}\geqslant−\mathrm{2},\:{y}\geqslant−\mathrm{3} \\ $$$${x}+\mathrm{2}−\mathrm{4}\sqrt{{x}+\mathrm{2}}+\mathrm{4}+{y}+\mathrm{3}−\mathrm{4}\sqrt{{y}+\mathrm{3}}+\mathrm{4}=\mathrm{13} \\ $$$$\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} +\left(\sqrt{{y}+\mathrm{3}}−\mathrm{2}\right)^{\mathrm{2}} =\left(\sqrt{\mathrm{13}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{2}}−\mathrm{2}=\sqrt{\mathrm{13}}\:\mathrm{cos}\:\theta \\ $$$$\sqrt{{y}+\mathrm{3}}−\mathrm{2}=\sqrt{\mathrm{13}}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{x}=\left(\mathrm{2}+\sqrt{\mathrm{13}}\:\mathrm{cos}\:\theta\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{2}+\mathrm{4}\sqrt{\mathrm{13}}\:\mathrm{cos}\:\theta+\mathrm{13}\:\mathrm{cos}^{\mathrm{2}} \:\theta \\ $$$$\Rightarrow{y}=\left(\mathrm{2}+\sqrt{\mathrm{13}}\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} −\mathrm{3}=\mathrm{1}+\mathrm{4}\sqrt{\mathrm{13}}\:\mathrm{sin}\:\theta+\mathrm{13}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$${x}+{y}=\mathrm{16}+\mathrm{4}\sqrt{\mathrm{13}}\left(\mathrm{cos}\:\theta+\mathrm{sin}\:\theta\right) \\ $$$${x}+{y}=\mathrm{16}+\mathrm{4}\sqrt{\mathrm{26}}\:\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{4}}\right) \\ $$$$\Rightarrow\left({x}+{y}\right)_{{max}} =\mathrm{16}+\mathrm{4}\sqrt{\mathrm{26}}\:\checkmark \\ $$$${at}\:{x}=−\mathrm{2}:\: \\ $$$$\left(\sqrt{{y}+\mathrm{3}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{13}−\left(−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\sqrt{{y}+\mathrm{3}}−\mathrm{2}=\mathrm{3} \\ $$$${y}=\mathrm{22} \\ $$$$\Rightarrow{x}+{y}=−\mathrm{2}+\mathrm{22}=\mathrm{20} \\ $$$${at}\:{y}=−\mathrm{3}: \\ $$$$\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{13}−\left(−\mathrm{2}\right)^{\mathrm{2}} =\mathrm{9} \\ $$$$\sqrt{{x}+\mathrm{2}}−\mathrm{2}=\mathrm{3} \\ $$$${x}=\mathrm{23} \\ $$$$\Rightarrow{x}+{y}=\mathrm{23}−\mathrm{3}=\mathrm{20} \\ $$$$\Rightarrow\left({x}+{y}\right)_{{min}} =\mathrm{20}\:\checkmark \\ $$
Answered by A5T last updated on 17/Mar/24
Another method to get (x+y)_(max) : ((a+b)/2)≥((((√a)+(√b))/2))^2 ⇒((x+2+y+3)/2)≥((((√(x+2))+(√(y+3)))/2))^2 ⇒(((x+y)/8))^2 ≤((x+y+5)/2); x+y=p⇒(p^2 /(32))≤p+5 ⇒p^2 −32p−160≤0⇒16−4(√(26))≤x+y≤4(√(26))+16 Equality when x+2=y+3⇒ x=1+y ⇒1+2y=+8(√(y+3)) ⇒y=2(√(26))+((15)/2)⇒x=2(√(26))+((17)/2) ⇒(x,y)=(2(√(26))+((17)/2),2(√(26))+((15)/2)) at x+y=4(√(26))+16
$${Another}\:{method}\:{to}\:{get}\:\left({x}+{y}\right)_{{max}} : \\ $$$$\frac{{a}+{b}}{\mathrm{2}}\geqslant\left(\frac{\sqrt{{a}}+\sqrt{{b}}}{\mathrm{2}}\right)^{\mathrm{2}} \Rightarrow\frac{{x}+\mathrm{2}+{y}+\mathrm{3}}{\mathrm{2}}\geqslant\left(\frac{\sqrt{{x}+\mathrm{2}}+\sqrt{{y}+\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{x}+{y}}{\mathrm{8}}\right)^{\mathrm{2}} \leqslant\frac{{x}+{y}+\mathrm{5}}{\mathrm{2}};\:{x}+{y}={p}\Rightarrow\frac{{p}^{\mathrm{2}} }{\mathrm{32}}\leqslant{p}+\mathrm{5} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\mathrm{32}{p}−\mathrm{160}\leqslant\mathrm{0}\Rightarrow\mathrm{16}−\mathrm{4}\sqrt{\mathrm{26}}\leqslant{x}+{y}\leqslant\mathrm{4}\sqrt{\mathrm{26}}+\mathrm{16} \\ $$$${Equality}\:{when}\:{x}+\mathrm{2}={y}+\mathrm{3}\Rightarrow\:{x}=\mathrm{1}+{y} \\ $$$$\Rightarrow\mathrm{1}+\mathrm{2}{y}=+\mathrm{8}\sqrt{{y}+\mathrm{3}}\:\Rightarrow{y}=\mathrm{2}\sqrt{\mathrm{26}}+\frac{\mathrm{15}}{\mathrm{2}}\Rightarrow{x}=\mathrm{2}\sqrt{\mathrm{26}}+\frac{\mathrm{17}}{\mathrm{2}} \\ $$$$\Rightarrow\left({x},{y}\right)=\left(\mathrm{2}\sqrt{\mathrm{26}}+\frac{\mathrm{17}}{\mathrm{2}},\mathrm{2}\sqrt{\mathrm{26}}+\frac{\mathrm{15}}{\mathrm{2}}\right)\:{at}\:{x}+{y}=\mathrm{4}\sqrt{\mathrm{26}}+\mathrm{16} \\ $$

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