# Find the maximum value of $\square OXPY$

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.

• How is point $P$ defined? If it's a fixed point, there are only a few choices for $X$ and $Y$. Feb 14, 2019 at 18:27
• @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 $P$ Feb 14, 2019 at 18:31
• You solution looks right except I think the maximum value is when $\theta=3\pi/4$. Check your calculations. Essentially you need to maximize $\sin \theta - \cos \theta$ Feb 14, 2019 at 18:53
• @Vasya Oh! Thanks, My calculation is wrong. Feb 14, 2019 at 18:56
• @Vasya Use identity $\sin(\theta)-\cos(\theta)=\sqrt(2)\sin(\theta -\frac{1}{4}\pi)=-\sqrt(2)\cos(\theta -\frac{3}{4}\pi)$ Feb 14, 2019 at 19:12