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

find minima of  (x_1 −x_2 )^2 +5+(√(1−(x_1 )^2 ))+(√(4x_2 ))  ∀ x_1 ,x_2 ∈R

$${find}\:{minima}\:{of} \\ $$$$\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{5}+\sqrt{\mathrm{1}−\left({x}_{\mathrm{1}} \right)^{\mathrm{2}} }+\sqrt{\mathrm{4}{x}_{\mathrm{2}} }\:\:\forall\:{x}_{\mathrm{1}} ,{x}_{\mathrm{2}} \in{R} \\ $$

Commented by MJS last updated on 02/Oct/19

−1≤x_1 ≤1 ⇒ x_1 =sin p  0≤x_2  ⇒ x_2 =q^2   f(p, q)=(q^2 −sin p)^2 +5+∣cos p∣+2q  it′s easy to see no maximum exists:  q → +∞ ⇒ f(p, q) → +∞  searching the minimum:  0≤(q^2 −sin p)^2        the minimum is at q^2 =sin p ⇔ p=2nπ±arcsin q^2   0≤∣cos p∣≤1       the minimum is at p=(n−(1/2))π  0≤2q       the minimum is at q=0  trying we find absolute minima at  f(0, 0)=f(±π, 0)=6  ⇒ minima at (x_1 =0∨x_1 =±1)∧x_2 =0

$$−\mathrm{1}\leqslant{x}_{\mathrm{1}} \leqslant\mathrm{1}\:\Rightarrow\:{x}_{\mathrm{1}} =\mathrm{sin}\:{p} \\ $$$$\mathrm{0}\leqslant{x}_{\mathrm{2}} \:\Rightarrow\:{x}_{\mathrm{2}} ={q}^{\mathrm{2}} \\ $$$${f}\left({p},\:{q}\right)=\left({q}^{\mathrm{2}} −\mathrm{sin}\:{p}\right)^{\mathrm{2}} +\mathrm{5}+\mid\mathrm{cos}\:{p}\mid+\mathrm{2}{q} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:\mathrm{no}\:\mathrm{maximum}\:\mathrm{exists}: \\ $$$${q}\:\rightarrow\:+\infty\:\Rightarrow\:{f}\left({p},\:{q}\right)\:\rightarrow\:+\infty \\ $$$$\mathrm{searching}\:\mathrm{the}\:\mathrm{minimum}: \\ $$$$\mathrm{0}\leqslant\left({q}^{\mathrm{2}} −\mathrm{sin}\:{p}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:{q}^{\mathrm{2}} =\mathrm{sin}\:{p}\:\Leftrightarrow\:{p}=\mathrm{2}{n}\pi\pm\mathrm{arcsin}\:{q}^{\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\mid\mathrm{cos}\:{p}\mid\leqslant\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:{p}=\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)\pi \\ $$$$\mathrm{0}\leqslant\mathrm{2}{q} \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{is}\:\mathrm{at}\:{q}=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{we}\:\mathrm{find}\:\mathrm{absolute}\:\mathrm{minima}\:\mathrm{at} \\ $$$${f}\left(\mathrm{0},\:\mathrm{0}\right)={f}\left(\pm\pi,\:\mathrm{0}\right)=\mathrm{6} \\ $$$$\Rightarrow\:\mathrm{minima}\:\mathrm{at}\:\left({x}_{\mathrm{1}} =\mathrm{0}\vee{x}_{\mathrm{1}} =\pm\mathrm{1}\right)\wedge{x}_{\mathrm{2}} =\mathrm{0} \\ $$

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