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Question-152829




Question Number 152829 by liberty last updated on 01/Sep/21
Answered by MJS_new last updated on 02/Sep/21
z^2 =1−x^2 −y^2   ⇒ f(x, y, z)=x^2 +y^2 +x+2y−1  (1) obviously x≥0∧y≥0 to get a maximum  (2) obviously x≤1∧y≤1  (3) x^2 +y^2 ≤1 ⇒ max (x^2 +y^2 ) =1 with y=(√(1−x^2 ))  ⇒ f(x, y, z)=x+2(√(1−x^2 ))  ⇒ f ′=1−((2x)/( (√(1−x^2 ))))=0 ⇒ x=((√5)/5) ⇒ y=((2(√5))/5) ⇒ z=0  ⇒ max (f(x, y, z)) =(√5)
$${z}^{\mathrm{2}} =\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$$\Rightarrow\:{f}\left({x},\:{y},\:{z}\right)={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{x}+\mathrm{2}{y}−\mathrm{1} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{obviously}\:{x}\geqslant\mathrm{0}\wedge{y}\geqslant\mathrm{0}\:\mathrm{to}\:\mathrm{get}\:\mathrm{a}\:\mathrm{maximum} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{obviously}\:{x}\leqslant\mathrm{1}\wedge{y}\leqslant\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant\mathrm{1}\:\Rightarrow\:\mathrm{max}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:=\mathrm{1}\:\mathrm{with}\:{y}=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{f}\left({x},\:{y},\:{z}\right)={x}+\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{f}\:'=\mathrm{1}−\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}=\mathrm{0}\:\Rightarrow\:{x}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\:\Rightarrow\:{y}=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\:\Rightarrow\:{z}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{max}\:\left({f}\left({x},\:{y},\:{z}\right)\right)\:=\sqrt{\mathrm{5}} \\ $$
Answered by mr W last updated on 02/Sep/21
such that f(x,y,z)=x+2y−z^2  is max.  ⇒z=0  ⇒x^2 +y^2 =1 ⇒x=cos θ, y=sin θ  f=cos θ+2sin θ=(√5) cos (θ−tan^(−1) 2)  f_(max) =(√5)  at θ=tan^(−1) 2 ⇒x=(1/( (√5))), y=(2/( (√5))), z=0
$${such}\:{that}\:{f}\left({x},{y},{z}\right)={x}+\mathrm{2}{y}−{z}^{\mathrm{2}} \:{is}\:{max}. \\ $$$$\Rightarrow{z}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{x}=\mathrm{cos}\:\theta,\:{y}=\mathrm{sin}\:\theta \\ $$$${f}=\mathrm{cos}\:\theta+\mathrm{2sin}\:\theta=\sqrt{\mathrm{5}}\:\mathrm{cos}\:\left(\theta−\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\right) \\ $$$${f}_{{max}} =\sqrt{\mathrm{5}} \\ $$$${at}\:\theta=\mathrm{tan}^{−\mathrm{1}} \mathrm{2}\:\Rightarrow{x}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}},\:{y}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}},\:{z}=\mathrm{0} \\ $$

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