MHB What Is the Maximum Value of \( x \) in the Equation \( 2x+y+\frac{4}{xy}=10 \)?

[anemone]

Here is this week's POTW:

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If the positive real numbers $x$ and $y$ satisfy the equation $2x+y+\dfrac{4}{xy}=10$, find the maximum possible value of $x$.

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[anemone]

No one answered last POTW correctly. Participants can refer to the suggested solution below:

Spoiler

From the given equation, we rewrite it to get

$5-x=\dfrac{y}{2}+\dfrac{2}{xy}>0$ and this implies $x<5$.

Squaring both sides we get

$(5-x)^2=\dfrac{y^2}{4}+\dfrac{2}{a}+\dfrac{4}{x^2y^2}=\dfrac{4}{a}+\left(\dfrac{y}{2}-\dfrac{2}{xy}\right)^2\ge\dfrac{4}{x}$

And so $x(x-5)^2\ge 4$, this translates to $(x-4)(x^2-6x+1)\ge 0$, which, considering $x<5$, implies $x\le 4$.

 

[bob012345]

Are we still allowed to post an answer for an old POTW?

 

[anemone]

Are we still allowed to post an answer for an old POTW?

Of course! I don't see why not, and as a matter of fact, thanks for showing interest in the old POTW!

 

[malawi_glenn]

Spoiler

Solve for $y$:
We get $y^2 + 2(x-5)y + \dfrac{4}{x}=0$
$y = 5-x \pm \sqrt{(x-5)^2- \dfrac{4}{x}}$
Now we study the discriminant $d(x) = (x-5)^2- \dfrac{4}{x}$
We have one root $x=4$, we then set $d(x) = 0$ and multiply both sides with $x$:
$0 = x^3-10x^2+25x - 4$
This third degree polynomial has $x=4$ as root, the other two roots we find by polymomial long-division:
$(x^3-10x^2+25x - 4)/(x-4) = x^2 - 6x+1$
Roots of $x^2 - 6x+1$ are $3-2\sqrt{2}$ and $3+2\sqrt{2}$.
Sketch of the discriminant $d(x) = (x-5)^2- \dfrac{4}{x}$ which we did using a sign-table:

Thus in the discriminant, we must have $3-2\sqrt{2} \leq x \leq 4$ or $x \geq 3+2\sqrt{2}$.
But since $x \geq 3+2\sqrt{2} > 5$ and $y = 5-x \pm \sqrt{(x-5)^2- \dfrac{4}{x}}$ this gives negative $y$.
Therefore, we must have that $3-2\sqrt{2} \leq x \leq 4$.

Answer: the largest $x$ value is 4.

 

[bob012345]

Here is my solution and a plot of the equation. I got the same range for $x$ as @malawi_glenn.

Spoiler

Here is my plot of the original equation formatted as $y= f(x)$, actually the sum of two solutions of the quadratic solution for $y$ and a function of $x$.

$$ 2x + y + \frac{4}{xy} =10$$
$$ y = \left( \frac{1}{2}\left(10 - 2x\right) +\frac{1}{2} \sqrt{(10-2x)^2 -\frac{16}{x} } \large \right)$$

$$ y = \left( \frac{1}{2}\left(10 - 2x\right) -\frac{1}{2} \sqrt{(10-2x)^2 -\frac{16}{x} } \large \right)$$
The green areas are where $y$ becomes a complex number and purple areas where $y$ is negative.

You can see the range of $y$ is exactly twice the range of $x$ as can be surmised from the symmetry of the equation.

 

[malawi_glenn]

@bob012345

Spoiler

Looks like a Dalitz plot :-)

 

[PeroK]

Another solution:

Spoiler

Multiply through by $xy$ and we have:
$$2x^2y + xy^2 + 4 - 10xy = 0$$Fix $x$ and let the left-hand side be $f(y)$.$$f'(y) = 2x^2 + 2xy - 10x = 2x(x + y - 5)$$And $f(y)$ has a minimum at $y = 5 - x$, where $$f_{min} = 2x^2(5-x) + x(5-x)^2 - 10x(5-x) + 4$$$$ = -x^3 + 10x^2 - 25x + 4 = -(x-4)(x^2 - 6x + 1)$$If $f_{min} > 0$, then there is no solution for $y$ for that $x$. We already have an obvious constraint from the original equation that $x < 5$, so we need $3 - 2\sqrt 2 \le x \le 4$. And we can check that $x = 4, y = 1$ is indeed a solution.

Alternatively, having noticed that $x = 4, y = 1$ is a solution and $x < 5$ we can see that $f_{min} > 0$ for $4 < x < 5$. Hence, there are no solutions for $x > 4$.