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Question Number 152829 by liberty last updated on 01/Sep/21

Answered by MJS_new last updated on 02/Sep/21

$$z^2=1-x^2-y^2$$$$\Rightarrow f(x,y,z)=x^2+y^2+x+2y-1$$$$\left(1\right)\mathrm{obviously}x⩾0\wedge y⩾0\mathrm{to}\mathrm{get}\mathrm{a}\mathrm{maximum}$$$$\left(2\right)\mathrm{obviously}x⩽1\wedge y⩽1$$$$\left(3\right)x^2+y^2⩽1\Rightarrow \max (x^2+y^2)=1\mathrm{with}y=\sqrt{1-x^2}$$$$\Rightarrow f(x,y,z)=x+2\sqrt{1-x^2}$$$$\Rightarrow f{}^{\prime }=1-\frac{2x}{\sqrt{1-x^2}}=0\Rightarrow x=\frac{\sqrt{5}}{5}\Rightarrow y=\frac{2\sqrt{5}}{5}\Rightarrow z=0$$$$\Rightarrow \max \left(f(x,y,z)\right)=\sqrt{5}$$

Answered by mr W last updated on 02/Sep/21

$$suchthatf(x,y,z)=x+2y-z^{2}ismax.$$$$\Rightarrow z=0$$$$\Rightarrow x^2+y^2=1\Rightarrow x=\cos \theta ,y=\sin \theta$$$$f=\cos \theta +2\sin \theta =\sqrt{5}\cos (\theta -{\tan }^{-1}2)$$$$f_{max}=\sqrt{5}$$$$at\theta ={\tan }^{-1}2\Rightarrow x=\frac{1}{\sqrt{5}},y=\frac{2}{\sqrt{5}},z=0$$

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