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Problem:

Let $x$ and $y$ be real numbers satisfying $\frac{x^2y^2 - 1}{2y-1}=3x.$ Find the largest possible value of $x.$

How would I do this? Would I multiply each side by $2y-1$?

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  • $\begingroup$ That might be a good start. Do you know about Lagrange multipliers? $\endgroup$ – Harald Hanche-Olsen May 7 '17 at 19:52
  • $\begingroup$ Never heard of them $\endgroup$ – JenkinsMa May 7 '17 at 19:54
  • $\begingroup$ Okay, then a direct method is called for. The answer by dxiv looks good. $\endgroup$ – Harald Hanche-Olsen May 7 '17 at 19:58
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Hint: eliminate the denominator and consider it as a quadratic equation in $y\,$: $$x^2 \cdot y^2 - 6x \cdot y +3x -1 = 0$$

For the quadratic to have real roots, its discriminant must be non-negative:

$$ \frac{1}{4} \Delta = 9 x^2 -x^2(3x -1) = x^2(10 - 3x) \ge 0 $$

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  • $\begingroup$ +1 out of curiosity, do you think implicit differentiation could solve this problem? $\endgroup$ – qbert May 7 '17 at 19:59
  • $\begingroup$ @qbert Don't know that it would, since the maximum is attained on the boundary of the region (but, even if it did, it would be an overkill in a case like this one). $\endgroup$ – dxiv May 7 '17 at 20:03
  • $\begingroup$ indeed, it is also difficult to tell when $x$ can be written as a function of $y$. Thanks! $\endgroup$ – qbert May 7 '17 at 20:05

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