The perimeter of a triangle is known to be $24$-units with one of the interior angles equal to $60^\circ $. Find the lengths of the three sides of the triangle if it has the maximum possible area.
Can the problem be done by method of one-variable calculus(i.e without using partial differentiation) or by some algebraic inequalities.
This proof uses a bit of topology and elementary knowledge of ellipses.
If the perimeter is $p=a+b+c$, then the set of potential lengths of the sides (for a potentially degenerate triangle!) is given by $a+b+c=p, a\ge 0, b\ge 0, c\ge 0, a\le b+c, b\le a+c, c\le a+b$, which is a compact subset of $\mathbb R^3$.
This means that the area (e.g. given by Heron's formula), as a continuous function of the lengths $a, b, c$, must achieve a maximum in the above set.
However, the maximum is obviously not achieved the triangle is degenerate. It is not achieved either if the triangle is not degenerate but two sides are nonequal. Namely, it is easy to see that, with a given $a$, which then fixes $b+c=p-a$, the maximum is achieved when $b=c$. (The locus of the corner opposite to side $a$ is an arc of an ellipse, and the point the most distant from side $a$ is on the bisector of this side.)
Thus, the maximum is achieved when all three sides are equal. This means $a=b=c=p/3$. This triangle will have an angle of $60^\circ$ and will therefore be the maximum among those triangles (having an angle of $60^\circ$).
Conclusion: the triangle is equilateral with sides $a=b=c=8$.