Given $x>0,y>0$, AND $\frac{3}{2 x^2+3 xy}+\frac{5}{3 xy+4 y^2} =2$, So max{xy}?

Question

Given $x>0;y>0$; if $\frac{3}{2 x^2+3 xy}+\frac{5}{3 xy+4 y^2} =2$, Find max{xy}?

Here is my try:

Solution 1:

$xy=t$,

$\frac{3}{2 x^2+3t}+\frac{5}{3t+4y^2} =2$

$y = \frac{t}{x}$

$\frac{3}{2 x^2+3t}+\frac{5}{3t+4(\frac{t}{x})^2} =2$

$f(x)=x^4(12t-10)+x^2(34t^2-24t)+(24t^3-12t^2)=0$

Is it possible to continue using my idea?

I know I can use another way to solve it:

Solution 2:

I tried to convert it to latex manually one letter by one letter and I can save some type operation with the help of this website:

https://www.simpletex.cn/ai/latex_ocr

Is there any better way to write it in latex?

I manually write it in the page within 6 minutes, but converting it to latex takes me over 20 minutes.

$$x y = x y \cdot \frac 1 2 ( \frac 3 { 2 x ^ 2+ 3 x y }+\frac5{3xy+4y^2})=\frac12(\frac{3y}{2x+3y}+\frac{5x}{3x+4y})$$

$$\begin{gathered}\begin{cases}2x+3y=m\\3x+8y=n\end{cases}\Rightarrow\begin{cases}6x+9y=3m\\6x+8y=2n\end{cases}\Rightarrow\begin{cases}x=3n-4m\\y=3m-2n\end{cases} \\xy=\frac{1}{2}(\frac{9m-6n}{m}+\frac{15n-20m}{n})=\frac{1}{2}(24-\frac{6n}{m}-\frac{20m}{n}) \end{gathered}$$

$$xy\le 12 - 2\sqrt{30}$$

Here is the reasoning:

easy to know: $$m > 0$$ $$n > 0$$

$$x=3n-4m > 0$$

$$y=3m-2n > 0$$

$$ \frac{4}{3}m < n < \frac{3}{2}m $$ i.e. $$ 1.33m < n < 1.5m $$

m = 2x + 3y > 0

n = 3x + 4y > 0

When the following equation is real, it will get the min.

$\frac{6}{}=\frac{20}{}$

At that time, $$ n = \frac{\sqrt{30}}{3}m = 1.82m $$

does not fulfill the above:

$$ 1.33m < n < 1.5m $$

So it will not occur.

Update to the answer 1:

I type in "\frac {1}{6}\left(5- \frac{13z+24}{6Power[z,2]+17z+12}\right) max" into the WolframAlpha. But it give me like the below

Answer 1

We're given $x \gt 0, y \gt 0 $ and

$ \dfrac{3}{2 x^2 + 3 xy } + \dfrac{5}{3 xy + 4 y^2} = 2 $

Multiply through by the common denominator,

$ 3 (3 xy + 4 y^2) + 5 (2 x^2 + 3 xy) = 2 (2 x^2 + 3 xy) (3 x y + 4 y^2 ) $

From which,

$ 10 x^2 + 12 y^2 + 24 xy = 2 x y ( 2 x + 3 y)( 3 x + 4 y ) $

So that,

$ 5 x^2 + 6 y^2 + 12 xy = x y ( 6 x^2 + 12 y^2 + 17 x y ) $

Now use polar coordinates, define

$ x= r \cos t$ , and $ y = r \sin t $ , where $ r \gt 0 $, then

the equation becomes

$ 5 \cos^2 t + 6 \sin^2 t + 12 \cos t \sin t \\ = r^2 \cos t \sin t (6 \cos^2 t + 12 \sin^2 t + 17 \cos t \sin t ) $

But $ x y = r^2 \cos t \sin t $ , therefore,

$ x y = \dfrac{ 5 \cos^2 t + 6 \sin^2 t + 12 \cos t \sin t }{6 \cos^2 t + 12 \sin^2 t + 17 \cos t \sin t} = \dfrac{ 5.5 - 0.5 \cos(2t) + 6 \sin(2t) }{9 - 3 \cos(2t) + 8.5 \sin(2 t) } $

Define the point in the $(U, V)$ plane as follows

$ U = 9 - 3 \cos(2t) + 8.5 \sin(2 t) $

$ V = 5.5 - 0.5 \cos(2t) + 6 \sin(2t) $

Note that $ 0 \lt t \lt \dfrac{\pi}{2} $

The point $(U,V)$ traces an ellipse in the $(U,V)$ plane, and that is plotted below. From the plot, it is clear that the maximum ratio $\dfrac{V}{U}$ occurs at $U = 6 , V = 5$. Therefore our maximum is $\dfrac{5}{6}$.

Answer 2

Setting $z=\frac xy$ you can rearrange the given equation to $$xy = \frac 16\left(5- \frac{13z+24}{6z^2+17z+12}\right)$$ So, there is no maximum but a supremum which is $\boxed{\frac 56}$.

Addendum due to comments:

We have $x,y > 0$. Hence, \begin{eqnarray*} \frac{3}{2 x^2+3 xy}+\frac{5}{3 xy+4 y^2} & = & 2 \\ & \Leftrightarrow & \\ \frac 12\left(\frac{3}{2\frac xy +3}+\frac{5}{3+4\frac yx}\right) & = & xy \\ & \stackrel{z=\frac xy > 0}{\Leftrightarrow} & \\ \frac 12\left(\frac{3}{2z +3}+\frac{5}{3+\frac 4z}\right) & = & xy \\ \end{eqnarray*} Now, you rearrange the LHS and do a polynomial division: \begin{eqnarray*} \frac 12\left(\frac{3}{2z +3}+\frac{5}{3+\frac 4z}\right) & = & \frac{5z^2+12z+6}{6z^2+17z+12} \\ & = & \frac 56 - \underbrace{\frac{\frac{13}6 z+4}{6z^2+17z+12}}_{q(z):= } \\ & \stackrel{z>0}{<} & \frac 56 \end{eqnarray*} Note that $\displaystyle \lim_{z\to\infty}q(z) = 0$.