Question Number 208553 by efronzo1 last updated on 18/Jun/24
$$\:\:\:\cancel{ } \\ $$
Answered by mr W last updated on 18/Jun/24
$${due}\:{to}\:{symmetry} \\ $$$${y}={x}\:\Rightarrow\mathrm{3}{x}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow{x}=\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${f}_{{min}} =−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${f}_{{max}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by efronzo1 last updated on 18/Jun/24
$$\mathrm{i}\:\mathrm{got}\:\mathrm{minimum}\:−\frac{\mathrm{17}}{\mathrm{8}} \\ $$
Commented by A5T last updated on 18/Jun/24
Answered by Frix last updated on 18/Jun/24
$${x}={u}−{v}\wedge{y}={u}+{v} \\ $$$${z}=\mathrm{2}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} +{u}\right) \\ $$$$\mathrm{3}{u}^{\mathrm{2}} +{v}^{\mathrm{2}} =\mathrm{1}\:\Rightarrow\:{v}^{\mathrm{2}} =\mathrm{1}−\mathrm{3}{u}^{\mathrm{2}} \\ $$$${z}=\mathrm{2}\left(\mathrm{4}{u}^{\mathrm{2}} +{u}−\mathrm{1}\right) \\ $$$${v}\in\mathbb{R}\:\Rightarrow\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\leqslant{u}\leqslant\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\Rightarrow\:\frac{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\leqslant{z}\leqslant\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\mathrm{BUT} \\ $$$${z}'=\mathrm{16}{u}+\mathrm{2}=\mathrm{0}\:\Rightarrow\:{u}=−\frac{\mathrm{1}}{\mathrm{8}}\:\wedge\:{z}''=\mathrm{16}>\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{min}\left({z}\right)=−\frac{\mathrm{17}}{\mathrm{8}} \\ $$$$\Rightarrow\:−\frac{\mathrm{17}}{\mathrm{8}}\leqslant{f}\left({x},\:{y}\right)\leqslant\frac{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$
Commented by efronzo1 last updated on 18/Jun/24
$$\mathrm{yes}\:\mathrm{min}\:\mathrm{value}\:\mathrm{is}\:−\frac{\mathrm{17}}{\mathrm{8}} \\ $$
Answered by A5T last updated on 18/Jun/24
![Suppose y=f(x) Then x^2 +y^2 +xy=1 ≡ x^2 +(f(x))^2 +xf(x)=1 (d/dx) of sides⇒2x+2f(x)f^′ (x)+xf^′ (x)+f(x)=0 ⇒f^′ (x)(2y+x)=−2x−y⇒f^′ (x)=((−2x−y)/(2y+x))...(i) Similarly,x+2xy+y=x+2xf(x)+f(x) (d/dx) of both sides⇒1+2xf′x+2f(x)+f^′ (x)=0 f^′ (x)=((−2y−1)/(2x+1))...(ii) (i)=(ii)⇒f^′ (x)=((−2y−1)/(2x+1))=((−2x−y)/(2y+x))[x≠((−1)/2),2y≠−x] ⇒4(x−y)(x+y)=−x+y=−(x−y) ⇒y=x or 4(x+y)=−1 y=x⇒3x^2 =1⇒x=+_− ((√3)/3)=y⇒f(x,y)=((2+_− (√3))/3) y=((−1)/4)−x⇒x=((+_− (√(61))−1)/8)⇒f(x,y)=((−17)/8) Comparing values⇒max(f(x,y))=((2+(√3))/3) and min(f(x,y))=((−17)/8)](https://www.tinkutara.com/question/Q208574.png)
$${Suppose}\:{y}={f}\left({x}\right) \\ $$$${Then}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{xy}=\mathrm{1}\:\equiv\:{x}^{\mathrm{2}} +\left({f}\left({x}\right)\right)^{\mathrm{2}} +{xf}\left({x}\right)=\mathrm{1} \\ $$$$\frac{{d}}{{dx}}\:{of}\:{sides}\Rightarrow\mathrm{2}{x}+\mathrm{2}{f}\left({x}\right){f}^{'} \left({x}\right)+{xf}^{'} \left({x}\right)+{f}\left({x}\right)=\mathrm{0} \\ $$$$\Rightarrow{f}^{'} \left({x}\right)\left(\mathrm{2}{y}+{x}\right)=−\mathrm{2}{x}−{y}\Rightarrow{f}^{'} \left({x}\right)=\frac{−\mathrm{2}{x}−{y}}{\mathrm{2}{y}+{x}}…\left({i}\right) \\ $$$$ \\ $$$${Similarly},{x}+\mathrm{2}{xy}+{y}={x}+\mathrm{2}{xf}\left({x}\right)+{f}\left({x}\right) \\ $$$$\frac{{d}}{{dx}}\:{of}\:{both}\:{sides}\Rightarrow\mathrm{1}+\mathrm{2}{xf}'{x}+\mathrm{2}{f}\left({x}\right)+{f}^{'} \left({x}\right)=\mathrm{0} \\ $$$${f}^{'} \left({x}\right)=\frac{−\mathrm{2}{y}−\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}…\left({ii}\right) \\ $$$$\left({i}\right)=\left({ii}\right)\Rightarrow{f}^{'} \left({x}\right)=\frac{−\mathrm{2}{y}−\mathrm{1}}{\mathrm{2}{x}+\mathrm{1}}=\frac{−\mathrm{2}{x}−{y}}{\mathrm{2}{y}+{x}}\left[{x}\neq\frac{−\mathrm{1}}{\mathrm{2}},\mathrm{2}{y}\neq−{x}\right] \\ $$$$\Rightarrow\mathrm{4}\left({x}−{y}\right)\left({x}+{y}\right)=−{x}+{y}=−\left({x}−{y}\right) \\ $$$$\Rightarrow{y}={x}\:{or}\:\mathrm{4}\left({x}+{y}\right)=−\mathrm{1} \\ $$$${y}={x}\Rightarrow\mathrm{3}{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow{x}=\underset{−} {+}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}={y}\Rightarrow{f}\left({x},{y}\right)=\frac{\mathrm{2}\underset{−} {+}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${y}=\frac{−\mathrm{1}}{\mathrm{4}}−{x}\Rightarrow{x}=\frac{\underset{−} {+}\sqrt{\mathrm{61}}−\mathrm{1}}{\mathrm{8}}\Rightarrow{f}\left({x},{y}\right)=\frac{−\mathrm{17}}{\mathrm{8}} \\ $$$${Comparing}\:{values}\Rightarrow{max}\left({f}\left({x},{y}\right)\right)=\frac{\mathrm{2}+\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${and}\:{min}\left({f}\left({x},{y}\right)\right)=\frac{−\mathrm{17}}{\mathrm{8}} \\ $$
Answered by efronzo1 last updated on 18/Jun/24
