MHB What Are the Maximum and Minimum Values of P Given the Equation?

[anemone]

Find the minimum and maximum of $P=\dfrac{y−x}{x+8y}$ for all real $x$ and $y$ that satisfy the equation $y^2(6-x^2)-xy-1=0$.

 

[anemone]

Hint:

Spoiler

If you're fond of using the AM-GM inequality method (when applicable) to solve for optimization problem, then you could try to express $P$ in terms of $xy$.

 

[Albert1]

Hint:

Spoiler

If you're fond of using the AM-GM inequality method (when applicable) to solve for optimization problem, then you could try to express $P$ in terms of $xy$.

Spoiler

$max(P)=\dfrac {7}{2},\,\, xy=-1$
$min(P)=\dfrac {-1}{10},\, xy=1$

 

[anemone]

Yes, those are the correct answers, Albert. If your method of solving is different than mine, I hope you could post it to share it with me and MHB. Thanks. :)

Here is my solution:

Spoiler

$\begin{align*}P&=\dfrac{y−x}{x+8y}\\&=\dfrac{y(y−x)}{y(x+8y)}\\&=\dfrac{y^2−xy}{xy+8y^2}\\&=\dfrac{\dfrac{x^2y^2+xy+1}{6}−xy}{xy+8\left(\dfrac{x^2y^2+xy+1}{6}\right)}\\&=\dfrac{x^2y^2+xy+1-6xy}{6xy+8x^2y^2+8xy+8}\\&=\dfrac{x^2y^2-5xy+1}{8x^2y^2+14xy+8}\\&=\dfrac{x^2y^2-5xy+1}{8(x^2y^2+\dfrac{14xy}{8}+1)}\\&=\dfrac{1}{8}\left(1-\dfrac{27xy}{4(x^2y^2+\dfrac{14xy}{8}+1)}\right)\\&=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}+xy+\dfrac{1}{xy})}\right)\end{align*}$

If $xy\gt 0$, we get $P_{\text{minimum}}=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}+2)}\right)=-\dfrac{1}{10}$.

If $xy\lt 0$, then we need to rewrite $P$ such that it takes the form:

$\begin{align*}P&=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}+xy+\dfrac{1}{xy})}\right)\\&=\dfrac{1}{8}\left(1-\dfrac{27}{4(\dfrac{14}{8}-|xy|-\dfrac{1}{|xy|})}\right)\\&=\dfrac{1}{8}\left(1+\dfrac{27}{4\left(|xy|+\dfrac{1}{|xy|}-\dfrac{14}{8}\right)}\right)\end{align*}$

Thus $P_{\text{maximum}}=\dfrac{1}{8}\left(1+\dfrac{27}{4\left(2-\dfrac{14}{8}\right)}\right)=\dfrac{7}{2}$.

 

[Albert1]

Spoiler

$max(P)=\dfrac {7}{2},\,\, xy=-1$
$min(P)=\dfrac {-1}{10},\, xy=1$

my solution:

Spoiler

let $k=xy$
$p=\dfrac {k^2-5k+1}{8k^2+14k+8}$ (up to now the solution is similar to anemone's method)
if $k>0 $ then :
$P\geq \dfrac{2k-5k}{8k^2+14k+8}=\dfrac {-3k}{8k^2+14k+8}=\dfrac {-1}{10}$ equality holds when $k=xy=1$
if $k<0 $ then :
$P\leq \dfrac{k^2-5k+1}{14k-16k}=\dfrac {k^2-5k+1}{-2k}=\dfrac {7}{2}$ equality holds when $k=xy=-1$