Question Number 151554 by mathdanisur last updated on 21/Aug/21
Answered by MJS_new last updated on 21/Aug/21
![f (x, y) =((x(x+y))/(4x^2 +xy+y^2 )) (d/dy)[f (x, y)]=0 ((x(3x^2 −2xy−y^2 ))/((4x^2 +xy+y^2 )^2 ))=0 x∈R^+ ⇔ x>0 ⇒ y=−3x∨y=x but x>0∧y>0 ⇒ y=x (d^2 /dy^2 )[f (x, x)]=−(1/(9x^2 ))<0 ⇒ maximum at y=x f (x, x) =(1/3) =max (f (x,y)) ⇒ max (f (x,y) +f (y, z) +f(z, x)) =1](https://www.tinkutara.com/question/Q151563.png)
$${f}\:\left({x},\:{y}\right)\:=\frac{{x}\left({x}+{y}\right)}{\mathrm{4}{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dy}}\left[{f}\:\left({x},\:{y}\right)\right]=\mathrm{0} \\ $$$$\frac{{x}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{xy}−{y}^{\mathrm{2}} \right)}{\left(\mathrm{4}{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} }=\mathrm{0} \\ $$$${x}\in\mathbb{R}^{+} \:\Leftrightarrow\:{x}>\mathrm{0} \\ $$$$\Rightarrow\:{y}=−\mathrm{3}{x}\vee{y}={x}\:\mathrm{but}\:{x}>\mathrm{0}\wedge{y}>\mathrm{0}\:\Rightarrow\:{y}={x} \\ $$$$\frac{{d}^{\mathrm{2}} }{{dy}^{\mathrm{2}} }\left[{f}\:\left({x},\:{x}\right)\right]=−\frac{\mathrm{1}}{\mathrm{9}{x}^{\mathrm{2}} }<\mathrm{0}\:\Rightarrow\:\mathrm{maximum}\:\mathrm{at}\:{y}={x} \\ $$$${f}\:\left({x},\:{x}\right)\:=\frac{\mathrm{1}}{\mathrm{3}}\:=\mathrm{max}\:\left({f}\:\left({x},{y}\right)\right) \\ $$$$\Rightarrow\:\mathrm{max}\:\left({f}\:\left({x},{y}\right)\:+{f}\:\left({y},\:{z}\right)\:+{f}\left({z},\:{x}\right)\right)\:=\mathrm{1} \\ $$
Commented by mathdanisur last updated on 22/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$