Problem: There are moving point $X$ and $Y$ lie on the $x$ and $y$ axes, respectively. For moving point $P$, $PX=3$ and $PY=4$. Find the maximum area of $\square OXPY$.($O$ is origin).

My solution:

$W.L.O.G$, $X=(a,0),Y=(0,b)$, $XY \leq 7$ ,then $0 \leq a^2+b^2 \leq 7^2$. By $A.M.$, $$a^2+b^2 \geq 2 |ab|$$ When $\theta=\angle YPX$ we get $$a^2+b^2=16+9-2\cdot3\cdot4 \cos(\theta)$$ Area of $\square OXPY$ is ${1 \over 2}ab +\triangle XPY$. So,$$\frac{1}{2}\cdot3\cdot4 \sin(\theta) + {25-2\cdot3\cdot4 \cos(\theta) \over 4} \geq A(\theta)$$. Maximum value is $$A(\theta) = 6\sin(\theta)-6\cos(\theta)+\frac{25}{4} ={25 \over 4}+6\sqrt2\sin(\theta-\frac{\pi}{4}) \leq {25 \over 4} + 6\sqrt{2}$$ , where $\theta=\frac{3}{4}\pi$

My intuition told me that

  • $\triangle XPY$ is maximum

  • $\triangle XPY$ is minimum

But both of them is $\frac{49}{4}$ and smaller than my answer.

Is my solution right? If it is right, why is my intuition wrong? And I would like to share if you know a different way of solution.

  • $\begingroup$ How is point $P$ defined? If it's a fixed point, there are only a few choices for $X$ and $Y$. $\endgroup$
    – Vasili
    Feb 14, 2019 at 18:27
  • 1
    $\begingroup$ @Vasya $P$ is moving point. But it's location relys on $X$ and $Y$. And there are two location when $XY \leq 7$, but $\square OPXY$ must be convex square so vertually, there are only one location for $P$ $\endgroup$
    – user366725
    Feb 14, 2019 at 18:31
  • $\begingroup$ You solution looks right except I think the maximum value is when $\theta=3\pi/4$. Check your calculations. Essentially you need to maximize $\sin \theta - \cos \theta$ $\endgroup$
    – Vasili
    Feb 14, 2019 at 18:53
  • $\begingroup$ @Vasya Oh! Thanks, My calculation is wrong. $\endgroup$
    – user366725
    Feb 14, 2019 at 18:56
  • $\begingroup$ @Vasya Use identity $\sin(\theta)-\cos(\theta)=\sqrt(2)\sin(\theta -\frac{1}{4}\pi)=-\sqrt(2)\cos(\theta -\frac{3}{4}\pi)$ $\endgroup$
    – user366725
    Feb 14, 2019 at 19:12


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