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if-x-2-xy-y-2-5-find-the-minimum-and-maximum-of-x-2-xy-y-2-




Question Number 177130 by mr W last updated on 01/Oct/22
if x^2 +xy+y^2 =5, find the minimum  and maximum of x^2 −xy+y^2 .
ifx2+xy+y2=5,findtheminimumandmaximumofx2xy+y2.
Commented by Frix last updated on 01/Oct/22
I think  (5/3)≤x^2 −xy+y^2 ≤15
Ithink53x2xy+y215
Commented by Frix last updated on 01/Oct/22
x^2 ±xy+y^2 =c  ⇒  (c/3)≤x^2 ∓xy+y^2 ≤3c
x2±xy+y2=cc3x2xy+y23c
Commented by mr W last updated on 01/Oct/22
thanks sir!
thankssir!
Commented by Frix last updated on 01/Oct/22
you are right, I don′t know what I did before...
youareright,IdontknowwhatIdidbefore
Answered by mr W last updated on 01/Oct/22
x^2 +xy+y^2 =c=5   ...(i)  let x^2 −xy+y^2 =k   ...(ii)  [(i)−(ii)]/2:  ⇒xy=((c−k)/2) ⇒y=((c−k)/(2x))  x^2 +((c−k)/2)+(((c−k)/(2x)))^2 =c  x^4 −(((c+k)/2))x^2 +(((c−k)/2))^2 =0  Δ=(((c+k)/2))^2 −4(((c−k)/2))^2 ≥0  (k−3c)(3k−c)≤0  ⇒(c/3)≤k≤3c  i.e. (c/3)≤ x^2 −xy+y^2 ≤3c  with c=5:  minimum=(5/3)  maximum=3×5=15
x2+xy+y2=c=5(i)letx2xy+y2=k(ii)[(i)(ii)]/2:xy=ck2y=ck2xx2+ck2+(ck2x)2=cx4(c+k2)x2+(ck2)2=0Δ=(c+k2)24(ck2)20(k3c)(3kc)0c3k3ci.e.c3x2xy+y23cwithc=5:minimum=53maximum=3×5=15
Commented by Strengthenchen last updated on 06/Oct/22
how can i to know use this  way to get  correcting answer if i never seem this form question before?
howcanitoknowusethiswaytogetcorrectinganswerifineverseemthisformquestionbefore?
Answered by mr W last updated on 02/Oct/22
an other method  let x=u+v, y=u−v  x^2 +xy+y^2 =c  (u+v)^2 +(u+v)(u−v)+(u−v)^2 =c  3u^2 +v^2 =c → elipse  ⇒u=(√(c/3)) cos θ, v=(√c) sin θ    k=x^2 −xy+y^2     =(u+v)^2 −(u+v)(u−v)+(u−v)^2     =u^2 +3v^2     =((c cos^2  θ)/3)+3c sin^2  θ    =((c cos^2  θ)/3)+3c(1−cos^2  θ)    =(3−((8 cos^2  θ)/3))c  ⇒k_(min) =(3−(8/3))c=(c/3)  ⇒k_(max) =(3−0)c=3c
\boldsymbolan\boldsymbolother\boldsymbolmethodletx=u+v,y=uvx2+xy+y2=c(u+v)2+(u+v)(uv)+(uv)2=c3u2+v2=celipseu=c3cosθ,v=csinθk=x2xy+y2=(u+v)2(u+v)(uv)+(uv)2=u2+3v2=ccos2θ3+3csin2θ=ccos2θ3+3c(1cos2θ)=(38cos2θ3)ckmin=(383)c=c3kmax=(30)c=3c

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