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Question Number 70167 by Kunal12588 last updated on 01/Oct/19

$$findminimaof$$$${(x_1-x_2)}^2+5+\sqrt{1-{\left(x_1\right)}^2}+\sqrt{4x_2}\mathrm{\forall }{x}_1,x_2\in R$$

Commented by MJS last updated on 02/Oct/19

$$-1⩽x_1⩽1\Rightarrow x_1=\sin p$$$$0⩽x_2\Rightarrow x_2=q^2$$$$f(p,q)={(q^2-\sin p)}^2+5+\mid \cos p\mid +2q$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{no}\mathrm{maximum}\mathrm{exists}:$$$$q\to +\mathrm{\infty }\Rightarrow f(p,q)\to +\mathrm{\infty }$$$$\mathrm{searching}\mathrm{the}\mathrm{minimum}:$$$$0⩽{(q^2-\sin p)}^2$$$$\mathrm{the}\mathrm{minimum}\mathrm{is}\mathrm{at}{q}^2=\sin p\iff p=2n\pi \pm \arcsin q^2$$$$0⩽\mid \cos p\mid ⩽1$$$$\mathrm{the}\mathrm{minimum}\mathrm{is}\mathrm{at}p=(n-\frac{1}{2})\pi$$$$0⩽2q$$$$\mathrm{the}\mathrm{minimum}\mathrm{is}\mathrm{at}q=0$$$$\mathrm{trying}\mathrm{we}\mathrm{find}\mathrm{absolute}\mathrm{minima}\mathrm{at}$$$$f(0,0)=f(\pm \pi ,0)=6$$$$\Rightarrow \mathrm{minima}\mathrm{at}(x_1=0\vee x_1=\pm 1)\wedge x_2=0$$

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