In $\triangle ABC$, height $h_a=2m$ and perimeter$=7m$, find maximum height $h_b$. In $\triangle ABC$, height $h_a=2m$ and perimeter$=7m$, find maximum height $h_b$ and write the formula used.
Note: Not easy or incomplete question, you don't have angles or any other data.
 A: A possible approach: assume that the projection of $A$ on the $BC$-line splits the $BC$-side in two segments having length $x$ and $y$. We have to fulfill
$$ x+y+\sqrt{x^2+4m^2}+\sqrt{y^2+4m^2} = 7m $$
and we have to maximize $h_b = \frac{2\Delta}{b}=\frac{2m(x+y)}{\sqrt{x^2+4m^2}}$. This problem can be tackled through Lagrange multipliers. However, it is probably easier to solve the dual problem, i.e. to find the minimum perimeter of a triangle with two given heights $h_a,h_b$.
It is also useful to recall that:
$$ \frac{p}{\Delta} = \frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}$$
$$ \frac{1}{\Delta}= \sqrt{\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\left(-\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\left(\frac{1}{h_a}-\frac{1}{h_b}+\frac{1}{h_c}\right)\left(\frac{1}{h_a}+\frac{1}{h_b}-\frac{1}{h_c}\right)}$$
This turns out to be quite a difficult problem. Have a look at this similar question (and my answer), too. The optimal $b$ is given by the stationary point of $b+7-\sqrt{b^2-4}$, that occurs at $b\approx 2.03737$.
