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Question Number 208976 by efronzo1 last updated on 30/Jun/24
Answered by mr W last updated on 30/Jun/24
say x^2 +y^2 =u, xy=v x^2 +y^2 ≥2(√(x^2 y^2 ))=2∣xy∣ ⇒u≥2∣v∣ x^2 +y^2 +xy=1 ⇒u+v=1 (x^2 +y^2 )xy+4=s, say ⇒uv=s−4 u, v are roots of z^2 −z+s−4=0 ⇒u,v=((1±(√(17−4s)))/2) ((1+(√(17−4s)))/2)≥2×((1−(√(17−4s)))/2) (√(17−4s))≥1 ⇒s≤((38)/9) → maximum ((1+(√(17−4s)))/2)≥−2×((1−(√(17−4s)))/2) (√(17−4s))≤3 ⇒s≥2 → minimum
$${say}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={u},\:{xy}={v} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} \geqslant\mathrm{2}\sqrt{{x}^{\mathrm{2}} {y}^{\mathrm{2}} }=\mathrm{2}\mid{xy}\mid \\ $$$$\Rightarrow{u}\geqslant\mathrm{2}\mid{v}\mid \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{1} \\ $$$$\Rightarrow{u}+{v}=\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){xy}+\mathrm{4}={s},\:{say} \\ $$$$\Rightarrow{uv}={s}−\mathrm{4} \\ $$$${u},\:{v}\:{are}\:{roots}\:{of}\:{z}^{\mathrm{2}} −{z}+{s}−\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow{u},{v}=\frac{\mathrm{1}\pm\sqrt{\mathrm{17}−\mathrm{4}{s}}}{\mathrm{2}} \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{17}−\mathrm{4}{s}}}{\mathrm{2}}\geqslant\mathrm{2}×\frac{\mathrm{1}−\sqrt{\mathrm{17}−\mathrm{4}{s}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{17}−\mathrm{4}{s}}\geqslant\mathrm{1} \\ $$$$\Rightarrow{s}\leqslant\frac{\mathrm{38}}{\mathrm{9}}\:\:\:\rightarrow\:{maximum} \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{17}−\mathrm{4}{s}}}{\mathrm{2}}\geqslant−\mathrm{2}×\frac{\mathrm{1}−\sqrt{\mathrm{17}−\mathrm{4}{s}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{17}−\mathrm{4}{s}}\leqslant\mathrm{3} \\ $$$$\Rightarrow{s}\geqslant\mathrm{2}\:\:\rightarrow\:{minimum} \\ $$
Answered by Frix last updated on 30/Jun/24
x^2 +xy+y^2 =1 is an ellipse with center ((0),(0) ) The extreme values of f(x, y)=x^3 y+xy^3 +4 occur at the vertices of the ellipse. Rotating by α=(π/4): x=(((√2)(u+v))/2)∧y=(((√2)(v−u))/2) ⇒ Ellipse: (u^2 /2)+((3v^2 )/2)=1 ⇔ v=±(1/( (√3)))(√(2−u^2 )) Vertices at (((±(√2))),(0) ) and ((0),((±((√6)/3))) ) f(u, v)=−(u^4 /2)+(v^4 /2)+4 ⇒ 2≤f≤((38)/9)
$${x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{1}\:\mathrm{is}\:\mathrm{an}\:\mathrm{ellipse}\:\mathrm{with}\:\mathrm{center}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{The}\:\mathrm{extreme}\:\mathrm{values}\:\mathrm{of}\:{f}\left({x},\:{y}\right)={x}^{\mathrm{3}} {y}+{xy}^{\mathrm{3}} +\mathrm{4}\:\mathrm{occur} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse}. \\ $$$$\mathrm{Rotating}\:\mathrm{by}\:\alpha=\frac{\pi}{\mathrm{4}}: \\ $$$${x}=\frac{\sqrt{\mathrm{2}}\left({u}+{v}\right)}{\mathrm{2}}\wedge{y}=\frac{\sqrt{\mathrm{2}}\left({v}−{u}\right)}{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\mathrm{Ellipse}:\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{3}{v}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{1}\:\Leftrightarrow\:{v}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\sqrt{\mathrm{2}−{u}^{\mathrm{2}} } \\ $$$$\mathrm{Vertices}\:\mathrm{at}\:\begin{pmatrix}{\pm\sqrt{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{and}\:\begin{pmatrix}{\mathrm{0}}\\{\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}}\end{pmatrix} \\ $$$${f}\left({u},\:{v}\right)=−\frac{{u}^{\mathrm{4}} }{\mathrm{2}}+\frac{{v}^{\mathrm{4}} }{\mathrm{2}}+\mathrm{4} \\ $$$$\Rightarrow\:\mathrm{2}\leqslant{f}\leqslant\frac{\mathrm{38}}{\mathrm{9}} \\ $$
Answered by dimentri last updated on 30/Jun/24
⊎2
$$\:\:\:\underline{\biguplus}\mathrm{2} \\ $$
Answered by A5T last updated on 30/Jun/24
Let y=f(x); x^2 +y^2 +xy=1⇒f′(x)=((−2x−y)/(x+2y)) (d/dx)(x^3 y+xy^3 +4)=0⇒f′(x)=((−3x^2 y−y^3 )/(x^3 +3xy^2 )) f′(x)=((−2x−y)/(x+2y))=((−3x^2 y−y^3 )/(x^3 +3xy^2 ))⇒x^4 +xy^3 =x^3 y+y^4 ⇒(x^2 −y^2 )(x^2 +y^2 )=xy(x^2 −y^2 ) ⇒x^2 =y^2 ...(I) or x^2 +y^2 =xy...(II) I:x^2 =y^2 ⇒y=+_− x⇒ { ((y=x⇒3x^2 =1⇒x=+_− (1/( (√3)))=y)),((y=−x⇒x^2 =1⇒x=+_− 1=+^− y)) :} II:x^2 +y^2 =xy⇒2xy=1 ⇒xy(x^2 +y^2 )+4=(xy)^2 +4=(1/4)+4=((17)/4) Testing (x,y)=(+_− (1/( (√3))),+_− (1/( (√3)))),(+_− 1,+^− 1) gives the maximum and minimum at (x,y)=(+_− (1/( (√3))),+_− (1/( (√3)))) and (+_− 1,+^− 1) respectively ⇒2≤x^3 y+xy^3 +4≤((38)/9)
$${Let}\:{y}={f}\left({x}\right);\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{1}\Rightarrow{f}'\left({x}\right)=\frac{−\mathrm{2}{x}−{y}}{{x}+\mathrm{2}{y}} \\ $$$$\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} {y}+{xy}^{\mathrm{3}} +\mathrm{4}\right)=\mathrm{0}\Rightarrow{f}'\left({x}\right)=\frac{−\mathrm{3}{x}^{\mathrm{2}} {y}−{y}^{\mathrm{3}} }{{x}^{\mathrm{3}} +\mathrm{3}{xy}^{\mathrm{2}} } \\ $$$${f}'\left({x}\right)=\frac{−\mathrm{2}{x}−{y}}{{x}+\mathrm{2}{y}}=\frac{−\mathrm{3}{x}^{\mathrm{2}} {y}−{y}^{\mathrm{3}} }{{x}^{\mathrm{3}} +\mathrm{3}{xy}^{\mathrm{2}} }\Rightarrow{x}^{\mathrm{4}} +{xy}^{\mathrm{3}} ={x}^{\mathrm{3}} {y}+{y}^{\mathrm{4}} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)={xy}\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={y}^{\mathrm{2}} …\left({I}\right)\:{or}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={xy}…\left({II}\right) \\ $$$${I}:{x}^{\mathrm{2}} ={y}^{\mathrm{2}} \Rightarrow{y}=\underset{−} {+}{x}\Rightarrow\begin{cases}{{y}={x}\Rightarrow\mathrm{3}{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\underset{−} {+}\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}={y}}\\{{y}=−{x}\Rightarrow{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\underset{−} {+}\mathrm{1}=\overset{−} {+}{y}}\end{cases} \\ $$$${II}:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={xy}\Rightarrow\mathrm{2}{xy}=\mathrm{1} \\ $$$$\Rightarrow{xy}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)+\mathrm{4}=\left({xy}\right)^{\mathrm{2}} +\mathrm{4}=\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{4}=\frac{\mathrm{17}}{\mathrm{4}} \\ $$$${Testing}\:\left({x},{y}\right)=\left(\underset{−} {+}\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},\underset{−} {+}\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right),\left(\underset{−} {+}\mathrm{1},\overset{−} {+}\mathrm{1}\right) \\ $$$${gives}\:{the}\:{maximum}\:{and}\:{minimum}\:{at}\: \\ $$$$\left({x},{y}\right)=\left(\underset{−} {+}\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},\underset{−} {+}\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:{and}\:\left(\underset{−} {+}\mathrm{1},\overset{−} {+}\mathrm{1}\right)\:{respectively} \\ $$$$\Rightarrow\mathrm{2}\leqslant{x}^{\mathrm{3}} {y}+{xy}^{\mathrm{3}} +\mathrm{4}\leqslant\frac{\mathrm{38}}{\mathrm{9}} \\ $$

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