Maximize area of a triangle with fixed perimeter If perimeter of a triangle is $2d$, what is the length of sides so the triangle has maximal area?  
I found some solution using circle and angles, but I think I have to use derivatives.  
I need help.

Let $a$, $b$ and $c$ be the sides of a triangle. The perimeter, $p=a+b+c$, is fixed and we want to find the values of $a$, $b$ and $c$ that give the triangle maximum area. Heron's formula says that the triangle's area is
$$A=\sqrt{s(s-a)(s-b)(a-c)}$$
where s is the semiperimeter $\frac{a+b+c}{2}=\frac{p}{2}.$
Because p is fixed, we can write $c=p-a-b$. Substituting this into the equation above and squaring we find that
\begin{eqnarray}
16A^2=p(p-2a)(p-2b)(2a+2b-p). \quad\quad(1)
\end{eqnarray}
In the first part, we fix $a$ and see what we can do with $b$ to get a maximum. To this end, differentiating with respect to $b$ gives
\begin{align*}
\nonumber 32A\frac{dA}{db}=p(p-2a)\left[(p-2b)(2)+(2a+2b-p)(-2)\right]  
\nonumber =4p(p-2a)(p-2b-a).  
\end{align*}
If we set this equal to $0$ to find the critical points we find there are two possibilities. The first is that $p=2a$ which leads to $a=\frac{p}{2}$ and $b=c=\frac{p}{4},$ which do not make a proper triangle.
The more interesting possibility is that $p-2b-a=0$, or that $b=\frac{p-a}{2}.$ The significance of this value for $b$ becomes apparent when we see that $c=p-a-b=p-a-\frac{p-a}{2}=\frac{p-a}{2}=b.$ Thus we have established that the triangle is at least iscosceles.  
In the second part, we use the value of $b$ just obtained, and see what we can do with $a$. Substituting for $b$ in (1) we find that
$$16A^2=p(p-2a)a^2$$
which we differentiate with respect to $a$ to get
$$32A\frac{dA}{da}=p\left[(p-2a)(2a)+a^2(-2)\right] = 2ap(p-3a).$$
Setting this equal to 0 again gives us $a=\frac{p}{3}.$
Substituting back, we find that $b = \frac{p-a}{2}=\frac{p}{3}$ and finally $c=p-a-b=\frac{p}{3}$ as well, proving that maximum area is achieved when the triangle is equilateral.
