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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)
$${if}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{5},\:{find}\:{the}\:{range}\:{of} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}. \\ $$$$\left({x},{y}\:\in\mathbb{R}\right) \\ $$
Commented by Frix last updated on 24/Oct/22
[(5/3); 15]
$$\left[\frac{\mathrm{5}}{\mathrm{3}};\:\mathrm{15}\right] \\ $$
Commented by Acem last updated on 25/Oct/22
Good!
$${Good}! \\ $$
Answered by Acem last updated on 25/Oct/22
R= [−5, +∞[
$$\:{R}=\:\left[−\mathrm{5},\:+\infty\left[\right.\right. \\ $$
Commented by MJS_new last updated on 25/Oct/22
how?
$$\mathrm{how}? \\ $$
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
$$\mathrm{let}\:{y}={px} \\ $$$$\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right){x}^{\mathrm{2}} =\mathrm{5} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{5}}{{p}^{\mathrm{2}} +{p}+\mathrm{1}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}=\frac{\mathrm{5}\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right)}{{p}^{\mathrm{2}} +{p}+\mathrm{1}} \\ $$$$\frac{{d}\left[\frac{\mathrm{5}\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right)}{{p}^{\mathrm{2}} +{p}+\mathrm{1}}\right]}{{dp}}=\mathrm{0} \\ $$$$\frac{\mathrm{10}\left({p}^{\mathrm{2}} −\mathrm{1}\right)}{\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow\:{p}=\pm\mathrm{1} \\ $$$$\Rightarrow \\ $$$$\frac{\mathrm{5}}{\mathrm{3}}\leqslant\frac{\mathrm{5}\left({p}^{\mathrm{2}} −{p}+\mathrm{1}\right)}{{p}^{\mathrm{2}} +{p}+\mathrm{1}}\leqslant\mathrm{15} \\ $$$$\Leftrightarrow \\ $$$$\frac{\mathrm{5}}{\mathrm{3}}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\leqslant\mathrm{15} \\ $$
Commented by Acem last updated on 25/Oct/22
You′re right
$$\:{You}'{re}\:{right} \\ $$
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
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{5}\:\:\:\:\:…\left({i}\right) \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}={k}\:\:\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{5}+{k}}{\mathrm{2}} \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${xy}=\frac{\mathrm{5}−{k}}{\mathrm{2}} \\ $$$$\frac{\mathrm{5}−{k}}{\mathrm{2}}={xy}\leqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}=\frac{\mathrm{5}+{k}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{10}−\mathrm{2}{k}\leqslant\mathrm{5}+{k}\:\Rightarrow{k}\geqslant\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\frac{\mathrm{5}−{k}}{\mathrm{2}}={xy}\geqslant−\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}=−\frac{\mathrm{5}+{k}}{\mathrm{4}} \\ $$$$\Rightarrow\mathrm{10}−\mathrm{2}{k}\geqslant−\mathrm{5}−{k}\:\Rightarrow{k}\leqslant\mathrm{15} \\ $$$$ \\ $$$$\Rightarrow\frac{\mathrm{5}}{\mathrm{3}}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{xy}\leqslant\mathrm{15} \\ $$
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)
$$\left({x}−{y}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}{xy}\:\Rightarrow{xy}\leqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} \geqslant\mathrm{0}\:\Rightarrow\mathrm{2}{xy}\geqslant−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:\Rightarrow{xy}\geqslant−\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}} \\ $$
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)
$${Nice}\:{solution},\:{but} \\ $$$$\frac{\mathrm{5}−{k}}{\mathrm{2}}={xy}\leqslant\:\frac{\mathrm{5}+{k}}{\mathrm{2}}\:{but}\:{how}\:{xy}\leqslant\:\frac{\mathrm{5}+{k}}{\mathrm{4}} \\ $$$$ \\ $$
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).
$${xy}\leqslant\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}}\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\frac{\mathrm{5}+{k}}{\mathrm{2}},\:{so}\:{it}\:{is} \\ $$$${clear}\:{to}\:{me}\:{that} \\ $$$${xy}\leqslant\frac{\mathrm{5}+{k}}{\mathrm{4}}. \\ $$
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)
$$ \\ $$$$ \\ $$$${find}\:\:\:\:\:{r}^{\mathrm{2}} \left(\mathrm{1}−\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right) \\ $$$${r}^{\mathrm{2}} \left(\:\mathrm{1}+\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\right)=\mathrm{5} \\ $$$${sin}\mathrm{2}\theta=\frac{\mathrm{10}}{{r}^{\mathrm{2}} }−\mathrm{2} \\ $$$$−\mathrm{1}\ll\frac{\mathrm{10}}{{r}^{\mathrm{2}} }−\mathrm{2}\leqslant\mathrm{1} \\ $$$$\frac{\mathrm{10}}{\mathrm{3}}\leqslant{r}^{\mathrm{2}} \leqslant\mathrm{10} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\leqslant\mathrm{1}−\frac{{sin}\mathrm{2}\theta}{\mathrm{2}}\leqslant\frac{\mathrm{3}}{\mathrm{2}} \\ $$

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