For real numbers $x, y$ satisfy the condition $x>0, y>0$ and $$\log _{\sqrt{3}}\left(\frac{x+y}{x^2+y^2+xy+2}\right)=x\left(x-3\right)+y\left(y-3\right)+xy$$ Find the largest value of the expression $$P=\frac{3x+2y+1}{x+y+6}$$


closed as off-topic by Aqua, qbert, eranreches, Nosrati, José Carlos Santos Dec 27 '17 at 18:28

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    $\begingroup$ there are a couple things wrong with the way this was asked in my opinion: 1) you put none of your thoughts and just copy pasted a problem you are working on 2) your tags make it unclear if you would accept a solution using calculus, as the one below does (and would certainly be the way I would expect it to be solved) $\endgroup$ – qbert Dec 27 '17 at 9:24
  • $\begingroup$ @Mary: You said you know a little english. Are you from Shanghai, China? $\endgroup$ – DeepSea Dec 27 '17 at 13:17
  • $\begingroup$ I am from Vietnam. $\endgroup$ – Mary Jan 3 '18 at 2:47

I think this is a very nice question.

We have $$P(x,y)=\frac{3x+2y+1}{x+y+6}\tag{1}$$ so by the Quotient Rule $$P'(x,y)=\frac{(3+2y')(x+y+6)-(3x+2y+1)(1+y')}{(x+y+6)^2}=\frac{y+(11-x)y'+17}{(x+y+6)^2}=0$$ for stationary points. Thus $$(11-x)y'+y+17=0\implies y'+\frac1{11-x}y=-\frac{17}{11-x}$$ is a first-order linear ODE, which is easy to solve.

Here $p(x)=\dfrac1{11-x}$ and $q(x)=-\dfrac{17}{11-x}$. The integrating factor is $$h(x)=\exp\left(\int p(x)\, dx\right)=\frac1{11-x}$$ so we solve $$y\cdot p(x)=\int p(x)q(x)\, dx \implies \frac y{11-x}=\int -\frac{17}{(11-x)^2}\, dx=-\frac{17}{11-x}+c$$ where $c$ is a constant. Hence $$y=-17-c(x-11)=-17+11c-cx\tag{2}$$ Plugging this back into $(1)$, we get $$\max P=\frac{3x-34+22c-2cx+1}{x-17+11c-cx+6}=\frac{(3-2c)(x-11)}{(1-c)(x-11)}=\frac{3-2c}{1-c}\tag{3}$$ Now let $x=0$. Then by $(2)$, we have $$y(0)=-17+11c\implies c=\frac1{11}(y(0)+17)\tag{4}$$ We are given the condition that $$\log _{\sqrt{3}}\left(\frac{x+y}{x^2+y^2+xy+2}\right)=x\left(x-3\right)+y\left(y-3\right)+xy$$ so $$\log_{\sqrt3}\left(\frac y{y^2+2}\right)=y(y-3)$$ This has solutions when $y=1, 2$.

Therefore, when $y=1$, $c=\dfrac{18}{11}$ from $(4)$ and so from $(3)$,$$P_{y=1}=\frac{3-2\left(\frac{18}{11}\right)}{1-\frac{18}{11}}=\frac37$$ and when $y=2$, $c=\dfrac{19}{11}$ from $(4)$ and so from $(3)$, $$P_{y=2}=\frac{3-2\left(\frac{19}{11}\right)}{1-\frac{19}{11}}=\frac58$$ Since $\dfrac58>\dfrac37$, $$\boxed{\max P = \frac58.}$$

  • $\begingroup$ why do you assume $y$ is a function of $x$ here? $\endgroup$ – qbert Dec 27 '17 at 9:26
  • $\begingroup$ @qbert I am not aware of any other method to solve this (since the condition has $\log$ on LHS and polynomials on the right - using the Lambert W function would be very complicated) Also, $y$ boils down to have a linear relationship with $x$. Is my answer wrong? $\endgroup$ – TheSimpliFire Dec 27 '17 at 9:31
  • $\begingroup$ I'm not sure, you could probably justify this with the implicit function theorem on the constraint condition. I am certainly not claiming it is wrong, I was just wondering if you had a justification $\endgroup$ – qbert Dec 27 '17 at 9:32

Unfortunately I'm in high school so I do not understand your solution at "a first-order linear ODE". According to my ability, from the condition of the problem I just exploited: ${{\log }_{\sqrt{3}}}\left( \frac{x+y}{{{x}^{2}}+{{y}^{2}}+xy+2} \right)=x\left( x-3 \right)+y\left( y-3 \right)+xy$ $\Leftrightarrow 2{{\log }_{3}}\left( x+y \right)-2{{\log }_{3}}\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)={{x}^{2}}+{{y}^{2}}+xy-3\left( x+y \right)$ $\Leftrightarrow 2{{\log }_{3}}3\left( x+y \right)+3\left( x+y \right)=\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)+2{{\log }_{3}}\left( {{x}^{2}}+{{y}^{2}}+xy+2 \right)$ $\Leftrightarrow 3\left( x+y \right)={{x}^{2}}+{{y}^{2}}+xy+2$

Please help me complete the problem solving on high school mathematics knowledge. (Sympathy for me because I know a little English so I have to use "Google translate" translation)


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