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.