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Question Number 179100 by mr W last updated on 24/Oct/22

if x^2 +y^2 +xy=5, find the range of  x^2 +y^2 −xy.  (x,y ∈R)

ifx2+y2+xy=5,findtherangeofx2+y2xy.(x,yR)

Commented by Frix last updated on 24/Oct/22

[(5/3); 15]

[53;15]

Commented by Acem last updated on 25/Oct/22

Good!

Good!

Answered by Acem last updated on 25/Oct/22

 R= [−5, +∞[

R=[5,+[

Commented by MJS_new last updated on 25/Oct/22

how?

how?

Answered by MJS_new last updated on 25/Oct/22

let y=px  (p^2 +p+1)x^2 =5  x^2 =(5/(p^2 +p+1))  x^2 +y^2 −xy=((5(p^2 −p+1))/(p^2 +p+1))  ((d[((5(p^2 −p+1))/(p^2 +p+1))])/dp)=0  ((10(p^2 −1))/((p^2 +p+1)^2 ))=0 ⇒ p=±1  ⇒  (5/3)≤((5(p^2 −p+1))/(p^2 +p+1))≤15  ⇔  (5/3)≤x^2 +y^2 −xy≤15

lety=px(p2+p+1)x2=5x2=5p2+p+1x2+y2xy=5(p2p+1)p2+p+1d[5(p2p+1)p2+p+1]dp=010(p21)(p2+p+1)2=0p=±1535(p2p+1)p2+p+11553x2+y2xy15

Commented by Acem last updated on 25/Oct/22

 You′re right

Youreright

Commented by mr W last updated on 25/Oct/22

yes, thanks!

yes,thanks!

Answered by mr W last updated on 25/Oct/22

x^2 +y^2 +xy=5     ...(i)  x^2 +y^2 −xy=k     ...(ii)  (i)+(ii):  x^2 +y^2 =((5+k)/2)  (i)−(ii):  xy=((5−k)/2)  ((5−k)/2)=xy≤((x^2 +y^2 )/2)=((5+k)/4)  ⇒10−2k≤5+k ⇒k≥(5/3)  ((5−k)/2)=xy≥−((x^2 +y^2 )/2)=−((5+k)/4)  ⇒10−2k≥−5−k ⇒k≤15    ⇒(5/3)≤x^2 +y^2 −xy≤15

x2+y2+xy=5...(i)x2+y2xy=k...(ii)(i)+(ii):x2+y2=5+k2(i)(ii):xy=5k25k2=xyx2+y22=5+k4102k5+kk535k2=xyx2+y22=5+k4102k5kk1553x2+y2xy15

Commented by mr W last updated on 25/Oct/22

(x−y)^2 ≥0 ⇒x^2 +y^2 ≥2xy ⇒xy≤((x^2 +y^2 )/2)  (x+y)^2 ≥0 ⇒2xy≥−(x^2 +y^2 ) ⇒xy≥−((x^2 +y^2 )/2)

(xy)20x2+y22xyxyx2+y22(x+y)202xy(x2+y2)xyx2+y22

Commented by Acem last updated on 25/Oct/22

Nice solution, but  ((5−k)/2)=xy≤ ((5+k)/2) but how xy≤ ((5+k)/4)

Nicesolution,but5k2=xy5+k2buthowxy5+k4

Commented by mr W last updated on 25/Oct/22

xy≤((x^2 +y^2 )/2) and x^2 +y^2 =((5+k)/2), so it is  clear to me that  xy≤((5+k)/4).

xyx2+y22andx2+y2=5+k2,soitiscleartomethatxy5+k4.

Commented by Acem last updated on 26/Oct/22

Good

Good

Answered by manxsol last updated on 26/Oct/22

    find     r^2 (1−((sin2θ)/2))  r^2 ( 1+((sin2θ)/2))=5  sin2θ=((10)/r^2 )−2  −1≪((10)/r^2 )−2≤1  ((10)/3)≤r^2 ≤10  (1/2)≤1−((sin2θ)/2)≤(3/2)

findr2(1sin2θ2)r2(1+sin2θ2)=5sin2θ=10r22110r221103r210121sin2θ232

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