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Question Number 151965 by mnjuly1970 last updated on 24/Aug/21
x , y ∈ R & sin(x )+ cos (y ) =1 then max ( sin(y) + cos (x) ) =? ....
$$ \\ $$$$\:\:\:{x}\:,\:{y}\:\in\:\mathbb{R}\: \\ $$$$\:\:\:\:\:\:\&\:{sin}\left({x}\:\right)+\:{cos}\:\left({y}\:\right)\:=\mathrm{1} \\ $$$${then}\:\:{max}\:\left(\:{sin}\left({y}\right)\:+\:{cos}\:\left({x}\right)\:\right)\:=? \\ $$$$\:\:…. \\ $$
Answered by mr W last updated on 24/Aug/21
let k=sin y+cos x ...(i) 1=sin x+cos y ...(ii) (i)^2 +(ii)^2 : k^2 +1^2 =1+1+2(sin ycos x+sin xcos y) k^2 =1+2 sin (x+y)≤1+2=3 ⇒k_(max) =(√3) ⇒k_(min) =−(√3)
$${let}\:{k}=\mathrm{sin}\:{y}+\mathrm{cos}\:{x}\:\:\:…\left({i}\right) \\ $$$$\mathrm{1}=\mathrm{sin}\:{x}+\mathrm{cos}\:{y}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)^{\mathrm{2}} +\left({ii}\right)^{\mathrm{2}} : \\ $$$${k}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} =\mathrm{1}+\mathrm{1}+\mathrm{2}\left(\mathrm{sin}\:{y}\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\mathrm{cos}\:{y}\right) \\ $$$${k}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}\:\mathrm{sin}\:\left({x}+{y}\right)\leqslant\mathrm{1}+\mathrm{2}=\mathrm{3} \\ $$$$\Rightarrow{k}_{{max}} =\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{k}_{{min}} =−\sqrt{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 24/Aug/21
very nice master W..gratefu<l..
$$\:\:{very}\:{nice}\:{master}\:{W}..{gratefu}<{l}.. \\ $$
Answered by john_santu last updated on 25/Aug/21
given x,y∈R and sin x+cos y=1 find max (sin y+cos x). by Langrange multiplier f(x,y,λ)=sin y+cos x+λ(sin x+cos y−1) (∂f/∂x)=−sin x+λcos x=0 ; tan x=λ (∂f/∂y)=cos y+λ(−sin y)=0 ; tan y=(1/λ) (∂f/∂λ)=sin x+cos y=1 (1)&(2)⇒tan x tan y=1 ⇒sin x sin y = cos x cos y ⇒(1−cos y)sin y=(√(1−(1−cos y)^2 )) cos y ⇒(1−cos y)(√(1−cos^2 y)) =(√(1−(1−cos y)^2 )) cos y ⇒(1−c)(√(1−c^2 )) = (√(2c−c^2 )) c ⇒(1−c)^2 (1−c^2 )=c^2 (2c−c^2 ) ⇒(1−2c+c^2 )(1−c^2 )=2c^3 −c^4 ⇒1−c^2 −2c+2c^3 +c^2 −c^4 =2c^3 −c^4 ⇒1−2c=0 ; c=(1/2)=cos y⇒sin y=± ((√3)/2) ⇒sin x=(1/2)⇒cos x=±((√3)/2) therefore max sin y+cos x = ((√3)/2)+((√3)/2)=(√3)
$$\:\mathrm{given}\:\mathrm{x},\mathrm{y}\in\mathbb{R}\:\mathrm{and}\:\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}=\mathrm{1} \\ $$$$\mathrm{find}\:\mathrm{max}\:\left(\mathrm{sin}\:\mathrm{y}+\mathrm{cos}\:\mathrm{x}\right). \\ $$$$\mathrm{by}\:\mathrm{Langrange}\:\mathrm{multiplier} \\ $$$$\mathrm{f}\left(\mathrm{x},\mathrm{y},\lambda\right)=\mathrm{sin}\:\mathrm{y}+\mathrm{cos}\:\mathrm{x}+\lambda\left(\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}−\mathrm{1}\right) \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{x}}=−\mathrm{sin}\:\mathrm{x}+\lambda\mathrm{cos}\:\mathrm{x}=\mathrm{0}\:;\:\mathrm{tan}\:\mathrm{x}=\lambda \\ $$$$\frac{\partial\mathrm{f}}{\partial\mathrm{y}}=\mathrm{cos}\:\mathrm{y}+\lambda\left(−\mathrm{sin}\:\mathrm{y}\right)=\mathrm{0}\:;\:\mathrm{tan}\:\mathrm{y}=\frac{\mathrm{1}}{\lambda} \\ $$$$\frac{\partial\mathrm{f}}{\partial\lambda}=\mathrm{sin}\:\mathrm{x}+\mathrm{cos}\:\mathrm{y}=\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\&\left(\mathrm{2}\right)\Rightarrow\mathrm{tan}\:\mathrm{x}\:\mathrm{tan}\:\mathrm{y}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{x}\:\mathrm{sin}\:\mathrm{y}\:=\:\mathrm{cos}\:\mathrm{x}\:\mathrm{cos}\:\mathrm{y} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{cos}\:\mathrm{y}\right)\mathrm{sin}\:\mathrm{y}=\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:\mathrm{y}\right)^{\mathrm{2}} }\:\mathrm{cos}\:\mathrm{y} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{cos}\:\mathrm{y}\right)\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{y}}\:=\sqrt{\mathrm{1}−\left(\mathrm{1}−\mathrm{cos}\:\mathrm{y}\right)^{\mathrm{2}} }\:\mathrm{cos}\:\mathrm{y} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{c}\right)\sqrt{\mathrm{1}−\mathrm{c}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{2c}−\mathrm{c}^{\mathrm{2}} }\:\mathrm{c} \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{c}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{c}^{\mathrm{2}} \right)=\mathrm{c}^{\mathrm{2}} \left(\mathrm{2c}−\mathrm{c}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\left(\mathrm{1}−\mathrm{2c}+\mathrm{c}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{c}^{\mathrm{2}} \right)=\mathrm{2c}^{\mathrm{3}} −\mathrm{c}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{c}^{\mathrm{2}} −\mathrm{2c}+\mathrm{2c}^{\mathrm{3}} +\mathrm{c}^{\mathrm{2}} −\mathrm{c}^{\mathrm{4}} =\mathrm{2c}^{\mathrm{3}} −\mathrm{c}^{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{2c}=\mathrm{0}\:;\:\mathrm{c}=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{cos}\:\mathrm{y}\Rightarrow\mathrm{sin}\:\mathrm{y}=\pm\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{cos}\:\mathrm{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{therefore}\:\mathrm{max}\:\mathrm{sin}\:\mathrm{y}+\mathrm{cos}\:\mathrm{x}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\sqrt{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 25/Aug/21
thank you so much...
$${thank}\:{you}\:{so}\:{much}… \\ $$

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