In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$ In $\triangle ABC$, if length of medians $BE$ and $CF$ are $12$ and $9$ respectively. Find $\triangle_{max}$
My attempt is as follows:-
$$\triangle=\dfrac{1}{2}bc\sin A$$
For having the maximum area,
$$A=90^{\circ}$$
$$\triangle_{max}=\dfrac{1}{2}bc$$

According to the above figure,
$$BE=12$$
$$AC=b$$
$$AE=EC=\dfrac{b}{2}$$
$$CF=9$$
$$BA=c$$
$$BF=FA=\dfrac{c}{2}$$
For $\triangle AEB$, 
$$AB^2+AE^2=BE^2$$
$$c^2+\left(\dfrac{b}{2}\right)^2=12^2$$
$$4c^2+b^2=144\cdot4\tag{1}$$
For $\triangle FAC$
$$AC^2+AF^2=CF^2$$
$$b^2+\left(\dfrac{c}{2}\right)^2=9^2$$
$$4b^2+c^2=81\cdot4\tag{2}$$
Solving equations $(1)$ and $(2)$
$$16b^2-b^2=81\cdot4\cdot4-144\cdot4$$
$$15b^2=144(9-4)$$
$$b^2=48$$
$$b=4\sqrt{3}$$
Putting the value of $b$ in $(1)$
$$4\cdot48+c^2=324$$
$$c^2=132$$
$$c=2\sqrt{33}$$
Hence $$\triangle_{max}=\dfrac{1}{2}8\sqrt{99}$$
$$\triangle_{max}=12\sqrt{11}$$
But actual answer is $72$. But this method seems to be correct, why am I not getting the correct answer from this method? Please help me in this.
 A: It's true that a triangle with given sides $b$ and $c$ attains it's maximum area when $\angle A = 90^{\circ}$.
But it's not true that a triangle with given medians attains it's maximum area when one of it's angle is a right angle which you assume in your attempt.
Here's one useful lemma to tackle the problem :

Given a triangle $\triangle ABC$ with area $S$ and median $m_a,m_b,m_c$. The area of triangle $\triangle XYZ$ whose sides lengths are equal to $m_a,m_b,m_c$ is equal to $\frac{3}{4}S$.

You could find the proof here or in Problems in Plane and Solid Geometry by V. Prasolov at page 26 problem 1.36.
Using this lemma, we construct another triangle $\triangle XYZ$ whose sides length area equal to the median lengths of $\triangle ABC$. Now, we are already given that two sides length of $\triangle XYZ$ are $12$ and $9$. Since $[ABC] = \frac{4}{3}[XYZ]$, in order to maximize $[ABC]$, it suffices to maximize $[XYZ]$. But clearly, the maximum area of $[XYZ]$ is $\frac{1}{2} \times 12 \times 9 = 54$. Thus, the maximum area of $ABC$ is $\frac{4}{3} \times 54 = 72$.
A: Let $m_{1}=12,m_{2}=9$ and $m_{3}$ be the third median.
Let $$s=\frac{m_{1}+m_{2}+m_{3}}{2}=\frac{21+m_{3}}{2}$$
So $$s-m_{1}=\frac{m_{3}-3}{2};s-m_{2}=\frac{m_{3}+3}{2};s-m_{3}=\frac{21-m_{3}}{2}$$
Now, area of triangle the length of whose medians are $m_{1},m_{2},m_{3}$ is given by (http://mathworld.wolfram.com/TriangleMedian.html)
$$\triangle=\frac{4}{3}\sqrt{s(s-m_{1})(s-m_{2})(s-m_{3})}=\frac{4}{3}\sqrt{2916-\frac{(m^2_{3}-225)^2}{16}}\leq 72$$
the maximum area being attained when $m_{3}=15$
