Finding the maximum area of a triangle when the distance from a point to three vertices is fixed (1).$P$ is inside $\Delta ABC$ and $$PA=2,PB=7,PC=11.$$Let the area of the triangle be $S$, then find The maximum value of $S$.
(2).$P$ is inside $\Delta ABC$ and $$PA=x,PB=y,PC=z.$$Let the area of the triangle be $S$, then find The maximum value of $S$.
My approach: for (1),considering the fixed $B,C$, we can easily know that the triangle area is largest when the $AP$ is perpendicular to $BC$. By symmetry, we can know that $P$ is pendicular.
Of course, this is not rigorous. How can we show that when the area reaches the maximum, $P$ must be pendicular without partial derivative.
 A: For the first case
$PA = 2, PB = 7 , PC = 11 $
So we let $P$ be at the origin $(0,0)$ , and we draw three circles $A,B,C$ of radii $2, 7, 11$ respectively.  Now we can select point $A$ to be at $(2, 0)$.
If point $B$ is fixed at $(x_2, y_2)$ , and we vary point $C$ along the perimeter of circle $C$, then to obtain the maximum area of $\triangle ABC$, point $C$ must have the maximum possible distance from the line segment $AB$, and this can only happen if the extension of segment $PC$ is perpendicular to segment $AB$, which is what you stated in the question.  Extending this result to all three vertices, we deduce that in the maximum area triangle, $PA$ is perpendicular to $BC$ , and $PB$ is perpendicular to $AC$ , and $PC$ is perpendicular to $AB$.  Hence point $P$ must be the orthocenter of $\triangle ABC$.
Now for the given values of radii, if $A$ is at $(2, 0)$, then $BC$ lies parallel to $y$ axis, i.e.
$B = (x, \sqrt{49 - x^2} )$
$ C = (x, -\sqrt{ 121 - x^2} )$
And we have to determine $x$ such that $B$ is perpendicular to $AC$
$ AC = (x - 2, - \sqrt{121 - x^2} ) $
Hence, by using the dot product
$ B \cdot AC = x (x - 2) - \sqrt{ (49 - x^2) (121 - x^2) } = 0 $
Hence,
$ x^2 - 2 x = \sqrt{ (49 - x^2)(121 - x^2) } $
Squaring
$ x^4 + 4 x^2 - 4 x^3 = (49)(121) - (170 x^2) + x^4 $
So that,
$ 4 x^3 - 174 x^2 + 5929 = 0 $
Solving gives the following solutions:
$-5.5 , 6.31346652052679 , 42.6865334794732 $
The valid root is the negative one: $-5.5$
Hence, $ BC = \sqrt{ 49 - (-5.5)^2 } + \sqrt{ 121 - (-5.5)^2 } = \sqrt{18.75} + \sqrt{90.75} $
Therefore, the maximum area is $ \frac{1}{2} (7.5) ( \sqrt{18.75} + \sqrt{90.75} ) = 30 \sqrt{3} $
