Smallest possible perimeter of a triangle if area is 135 What is the smallest possible perimeter of a (flat, regular) triangle if the area is 135?
I tried various equations but I always ended up with two unknown variables.
 A: The following proof is more complicated than the calculus approach. However, it uses some interesting ideas.
Our problem is equivalent to showing that among all triangles with a given perimeter, the equilateral triangle has maximum area. 
Let the perimeter be $p$, and let $s=p/2$ be the half-perimeter. If the sides are $a$, $b$, and $c$, then $a+b+c=2s$. By Heron's Formula, the area of the triangle is
$$\sqrt{s(s-a)(s-b)(s-c)}.$$
We want to maximize area, so equivalently we want to maximize $s(s-a)(s-b)(s-c)$. But since the perimeter is given, so constant, we want to maximize
$(s-a)(s-b)(s-c)$ under the constraint $a+b+c=2s$.
Now we quote the case $n=3$ of the important Arithmetic Mean - Geometric Mean Inequality known as AM-GM to high school students who are serious about math contests. 
Let $x$, $y$, and $z$ be positive. Then 
$$\frac{x+y+z}{3} \ge \sqrt[3]{xyz},$$
with equality only when $x=y=z$.
Let $x=s-a$, $y=s-b$, and $z=s-c$. Then $x+y+z=s$, and therefore
$$\frac{s}{3} \ge  \sqrt[3]{(s-a)(s-b)(s-c)},$$
with equality precisely when $s-a=s-b=s-c$, meaning that $a=b=c$.
Thus
$$\sqrt{s(s-a)(s-b)(s-c)}\le \sqrt{s(s^3/27)}=\frac{s^2}{3\sqrt{3}},$$ 
with equality only for the equilateral triangle with perimeter $2s$. 
A: Hint: You want an equilateral triangle.  That should give you a single equation.
