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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
findminimaof(x1x2)2+5+1(x1)2+4x2x1,x2R
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
1x11x1=sinp0x2x2=q2f(p,q)=(q2sinp)2+5+cosp+2qitseasytoseenomaximumexists:q+f(p,q)+searchingtheminimum:0(q2sinp)2theminimumisatq2=sinpp=2nπ±arcsinq20⩽∣cosp∣⩽1theminimumisatp=(n12)π02qtheminimumisatq=0tryingwefindabsoluteminimaatf(0,0)=f(±π,0)=6minimaat(x1=0x1=±1)x2=0

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