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.

Here is another way. Suppose that the lengths of the sides of the triangle are $a$,$b$ and $c$ such that that the perimeter of the triangle is fixed and it is $2s$. Using the Heron's formula, the area, $A$ is $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ The AM-GM inequality for three positive reals $a,b,c$ states that $\displaystyle (abc)^\frac{1}{3}\leq \frac{a+b+c}{3} $wih equality at $a=b=c$.We may use it above on $A$.

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.

Area is $\sqrt{s* (s-a)* (s-b)* (s-c)}$

if s is (a+b+c)/2

since we are trying to maximize the Area, we could say we are trying to maximize the area squared so just maximize

s * (s-a) * (s-b) * (s-c)

since s is a constant because the perimeter is a fixed quantity, this is equivalent to maximizing (s-a) * (s-b) * (s-c)

but also note (s-a) + (s-b) + (s-c) = 3s - perimeter = s

which is a constant

so, by AM-GM where xyz is the quantity, we are maximizing and x+y+z is a constant x=y=z

so, (s-a)=(s-b)=(s-c)

so, a=b=c

so, the triangle is equilateral