Finding the largest area of a right-angled triangle using Lagrange multipliers The area of a triangle of length a, b and c is given by $$A =\sqrt{s(s-a)(s-b)(s-c)}$$ where its perimeter is $2s$ such that $2s = a + b + c$
Consider a right-angled triangle with hypotenuse $a$ such that $a^2 = b^2 + c^2$. Using Lagrange multipliers find the largest area of a right angled triangle of perimeter $2s$ and express your answer in terms of the perimeter (i.e. in terms of $s$)
From this I understand that we have two constraints of $2s = a + b + c$ and $a^2 = b^2 + c^2$ on the equation $\sqrt{s(s-a)(s-b)(s-c)}$. 
Therefore I constructed using Lagrange multipliers that $$L_a = \frac{-s(s-b)(s-c)}{2p} + \lambda - 2\mu a = 0$$ $$L_b = \frac{-s(s-a)(s-c)}{2p} + \lambda + 2\mu b = 0$$ $$L_c = \frac{-s(s-a)(s-b)}{2p} + \lambda + 2\mu c = 0$$ $$L_\lambda = 2s - a - b - c = 0 $$ $$L_\mu = a^2 - b^2 - c^2 = 0$$ 
where $p=\sqrt{s(s-a)(s-b)(s-c)}$ , $\lambda$ and $\mu$ being arbitrary constants. 
I have attempted to solve this by multiplying $L_a$ by $(s-a)$, $L_b$ by $(s-b)$ and $L_c$ by $(s-c)$, adding them together and using the two constraints ($L_\lambda$ and $L_\mu$) to solve for $p$ (which represents the largest area). But it didn't seem to work. 
What I am stuck on here is I'm not sure how I can solve for $p$ in terms of $s$ using the 5 equations above. 
Thank you very much for your help.

We have:
$$0 = L_a - L_b = -\frac{s(s-c)(b-a)}{2p} - 2\mu(a + b)$$
$$0 = L_b - L_c = -\frac{s(s-a)(c-b)}{2p} + 2\mu(b- c)$$
Multiply the first equation by $(b-c)$, the second one by $(a + b)$ and add them:
\begin{align}
0 &= -\frac{s}{2p}\Big((s-c)(b-a)(b-c) + (s-a)(c-b)(a+b)\Big)\\
&= -\frac{s(b-c)}{2p}\Big((s-c)(b-a)- (s-a)(a+b)\Big)\\
&= -\frac{s(b-c)}{4p}\Big((a+b-c)(b-a)- (b+c-a)(a+b)\Big)\\
&=-\frac{s(b-c)}{4p}\Big(-2bc\Big)\\
&=\frac{s(b-c)bc}{2p}\\
\end{align}
The only term which can be zero is $b - c$ so we conclude $b = c$.
Now we have $$a^2 = b^2 + c^2 = 2b^2 \implies a = \sqrt{2}b$$
Then $$2s = a + b + c = (2 + \sqrt{2})b \implies b = \frac{2s}{2 + \sqrt2}$$
Finally we get
$$p = \frac{bc}{2} = \frac{b^2}{2} = \frac{2s^2}{6 + 4\sqrt{2}}$$
