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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 .
$${if}\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{5},\:{find}\:{the}\:{minimum} \\ $$$${and}\:{maximum}\:{of}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} . \\ $$
Commented by Frix last updated on 01/Oct/22
I think  (5/3)≤x^2 −xy+y^2 ≤15
$$\mathrm{I}\:\mathrm{think} \\ $$$$\frac{\mathrm{5}}{\mathrm{3}}\leqslant{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \leqslant\mathrm{15} \\ $$
Commented by Frix last updated on 01/Oct/22
x^2 ±xy+y^2 =c  ⇒  (c/3)≤x^2 ∓xy+y^2 ≤3c
$${x}^{\mathrm{2}} \pm{xy}+{y}^{\mathrm{2}} ={c} \\ $$$$\Rightarrow \\ $$$$\frac{{c}}{\mathrm{3}}\leqslant{x}^{\mathrm{2}} \mp{xy}+{y}^{\mathrm{2}} \leqslant\mathrm{3}{c} \\ $$
Commented by mr W last updated on 01/Oct/22
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by Frix last updated on 01/Oct/22
you are right, I don′t know what I did before...
$$\mathrm{you}\:\mathrm{are}\:\mathrm{right},\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{what}\:\mathrm{I}\:\mathrm{did}\:\mathrm{before}… \\ $$
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
$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={c}=\mathrm{5}\:\:\:…\left({i}\right) \\ $$$${let}\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} ={k}\:\:\:…\left({ii}\right) \\ $$$$\left[\left({i}\right)−\left({ii}\right)\right]/\mathrm{2}: \\ $$$$\Rightarrow{xy}=\frac{{c}−{k}}{\mathrm{2}}\:\Rightarrow{y}=\frac{{c}−{k}}{\mathrm{2}{x}} \\ $$$${x}^{\mathrm{2}} +\frac{{c}−{k}}{\mathrm{2}}+\left(\frac{{c}−{k}}{\mathrm{2}{x}}\right)^{\mathrm{2}} ={c} \\ $$$${x}^{\mathrm{4}} −\left(\frac{{c}+{k}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\left(\frac{{c}−{k}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\frac{{c}+{k}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{{c}−{k}}{\mathrm{2}}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\ $$$$\left({k}−\mathrm{3}{c}\right)\left(\mathrm{3}{k}−{c}\right)\leqslant\mathrm{0} \\ $$$$\Rightarrow\frac{{c}}{\mathrm{3}}\leqslant{k}\leqslant\mathrm{3}{c} \\ $$$${i}.{e}.\:\frac{{c}}{\mathrm{3}}\leqslant\:{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \leqslant\mathrm{3}{c} \\ $$$${with}\:{c}=\mathrm{5}: \\ $$$${minimum}=\frac{\mathrm{5}}{\mathrm{3}} \\ $$$${maximum}=\mathrm{3}×\mathrm{5}=\mathrm{15} \\ $$
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?
$${how}\:{can}\:{i}\:{to}\:{know}\:{use}\:{this}\:\:{way}\:{to}\:{get}\:\:{correcting}\:{answer}\:{if}\:{i}\:{never}\:{seem}\:{this}\:{form}\:{question}\:{before}? \\ $$
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
$$\boldsymbol{{an}}\:\boldsymbol{{other}}\:\boldsymbol{{method}} \\ $$$${let}\:{x}={u}+{v},\:{y}={u}−{v} \\ $$$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} ={c} \\ $$$$\left({u}+{v}\right)^{\mathrm{2}} +\left({u}+{v}\right)\left({u}−{v}\right)+\left({u}−{v}\right)^{\mathrm{2}} ={c} \\ $$$$\mathrm{3}{u}^{\mathrm{2}} +{v}^{\mathrm{2}} ={c}\:\rightarrow\:{elipse} \\ $$$$\Rightarrow{u}=\sqrt{\frac{{c}}{\mathrm{3}}}\:\mathrm{cos}\:\theta,\:{v}=\sqrt{{c}}\:\mathrm{sin}\:\theta \\ $$$$ \\ $$$${k}={x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \\ $$$$\:\:=\left({u}+{v}\right)^{\mathrm{2}} −\left({u}+{v}\right)\left({u}−{v}\right)+\left({u}−{v}\right)^{\mathrm{2}} \\ $$$$\:\:={u}^{\mathrm{2}} +\mathrm{3}{v}^{\mathrm{2}} \\ $$$$\:\:=\frac{{c}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{3}}+\mathrm{3}{c}\:\mathrm{sin}^{\mathrm{2}} \:\theta \\ $$$$\:\:=\frac{{c}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{3}}+\mathrm{3}{c}\left(\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\theta\right) \\ $$$$\:\:=\left(\mathrm{3}−\frac{\mathrm{8}\:\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{3}}\right){c} \\ $$$$\Rightarrow{k}_{{min}} =\left(\mathrm{3}−\frac{\mathrm{8}}{\mathrm{3}}\right){c}=\frac{{c}}{\mathrm{3}} \\ $$$$\Rightarrow{k}_{{max}} =\left(\mathrm{3}−\mathrm{0}\right){c}=\mathrm{3}{c} \\ $$

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