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Question Number 169397 by mokys last updated on 29/Apr/22

Answered by mr W last updated on 30/Apr/22

AC≤AB+BC (√((x+y)^2 +(3+1)^2 ))≤(√(x^2 +3^2 ))+(√(y^2 +1^2 ))=5 (x+y)^2 +(3+1)^2 ≤25 (x+y)^2 ≤9 −3≤x+y≤3 (x+y)_(max) =3 ✓

$${AC}\leqslant{AB}+{BC} \\ $$$$\sqrt{\left({x}+{y}\right)^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{1}\right)^{\mathrm{2}} }\leqslant\sqrt{{x}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\mathrm{5} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} +\left(\mathrm{3}+\mathrm{1}\right)^{\mathrm{2}} \leqslant\mathrm{25} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} \leqslant\mathrm{9} \\ $$$$−\mathrm{3}\leqslant{x}+{y}\leqslant\mathrm{3} \\ $$$$\left({x}+{y}\right)_{{max}} =\mathrm{3}\:\checkmark \\ $$

Commented by mr W last updated on 30/Apr/22

Answered by cortano1 last updated on 30/Apr/22

f(x,y,λ)=x+y+λ((√(x^2 +9)) +(√(y^2 +1))−5) f_x ′=1+λ((x/( (√(x^2 +9)))))=0 f_y ′=1+λ((y/( (√(y^2 +1)))))=0 f_λ ′= (√(x^2 +9))+(√(y^2 +1))−5=0 ⇒λ = λ ; ((√(x^2 +9))/x) = ((√(y^2 +1))/y) ⇒((x^2 +9)/x^2 ) = ((y^2 +1)/y^2 ) ; 1+(9/x^2 ) = 1+(1/y^2 ) ⇒x^2 = 9y^2 ⇒ { ((x=3y)),((x=−3y)) :}; (√(9y^2 +9)) +(√(y^2 +1)) = 5 ⇒9y^2 +9 = 26+y^2 −10(√(y^2 +1)) ⇒10(√(y^2 +1)) = 17−8y^2 ⇒100y^2 +100=64y^4 −272y^2 +289 ⇒ { ((y=± (3/4))),((y=± ((√(21))/2) (not solution))) :} ⇒x+y= { ((3y+y=4y=± 3)),((−3y+y=−2y=∓(3/2))) :} max = 3

$${f}\left({x},{y},\lambda\right)={x}+{y}+\lambda\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}\:+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}−\mathrm{5}\right) \\ $$$$\:{f}_{{x}} '=\mathrm{1}+\lambda\left(\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}\right)=\mathrm{0} \\ $$$$\:{f}_{{y}} '=\mathrm{1}+\lambda\left(\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}\right)=\mathrm{0} \\ $$$$\:{f}_{\lambda} '=\:\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\lambda\:=\:\lambda\:;\:\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{9}}}{{x}}\:=\:\frac{\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}}{{y}} \\ $$$$\Rightarrow\frac{{x}^{\mathrm{2}} +\mathrm{9}}{{x}^{\mathrm{2}} }\:=\:\frac{{y}^{\mathrm{2}} +\mathrm{1}}{{y}^{\mathrm{2}} }\:;\:\mathrm{1}+\frac{\mathrm{9}}{{x}^{\mathrm{2}} }\:=\:\mathrm{1}+\frac{\mathrm{1}}{{y}^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\:\mathrm{9}{y}^{\mathrm{2}} \:\Rightarrow\begin{cases}{{x}=\mathrm{3}{y}}\\{{x}=−\mathrm{3}{y}}\end{cases};\:\sqrt{\mathrm{9}{y}^{\mathrm{2}} +\mathrm{9}}\:+\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{5} \\ $$$$\Rightarrow\mathrm{9}{y}^{\mathrm{2}} +\mathrm{9}\:=\:\mathrm{26}+{y}^{\mathrm{2}} −\mathrm{10}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow\mathrm{10}\sqrt{{y}^{\mathrm{2}} +\mathrm{1}}\:=\:\mathrm{17}−\mathrm{8}{y}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{100}{y}^{\mathrm{2}} +\mathrm{100}=\mathrm{64}{y}^{\mathrm{4}} −\mathrm{272}{y}^{\mathrm{2}} +\mathrm{289} \\ $$$$\Rightarrow\begin{cases}{{y}=\pm\:\frac{\mathrm{3}}{\mathrm{4}}}\\{{y}=\pm\:\frac{\sqrt{\mathrm{21}}}{\mathrm{2}}\:\left({not}\:{solution}\right)}\end{cases} \\ $$$$\Rightarrow{x}+{y}=\:\begin{cases}{\mathrm{3}{y}+{y}=\mathrm{4}{y}=\pm\:\mathrm{3}}\\{−\mathrm{3}{y}+{y}=−\mathrm{2}{y}=\mp\frac{\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$$\:{max}\:=\:\mathrm{3} \\ $$

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