Use the mathod of Lagrange multipliers to prove that of all

Solution

Let a, b, c be the sides of the triangle and let p be its semiperimeter. Then, according to a well known formula (Heron\'s formula, if I\'m not mistaken), the triangle area is A = sqrt(p(p - a)(p - b)(p - c)) 1) p is fixed and we want to maxime A = sqrt(p(p - a)(p - b)(p - c)) subject to a + b + c = 2p Since u ? u² is an strictly increasing function for u = 0 and p > 0 is constant, this is equivalent to maximizing (p - a)(p - b)(p - c)) subject to a + b + c = 2p Since the cube root function is strictly increasing, this, in turn, is equivalent to maximizing [(p - a)(p - b)(p - c))]^(1/3) subject to a + b + c = 2p Now, with a simple algebraic transformation in the constraint equation, we see this is the same as maximizing [(p - a)(p - b)(p - c))]^(1/3) subject to [(p -a) + (p - b ) + (p-c)]/3 = p/3 Hence, we want to maximize the geometric mean of (p -a), (p - b) and (p - c) (3 positive numbers) with their arithmetic mean fixed in p/3. According to the arithmetic/geometric means inequalty, this will be achieved if, and only if, p -a = p - b = p -c, that is, if and only if, a = b = c (other wise, AM > GM). And we conclude the area is maximum when the triangle is equilateral. 2) is just the dual of 1 (I guess you mean the minimum perimeter, of course. There\'s a typo in your statement) The same reasoning proves it, considering that now we want to minimize the perimeter for fixed A. We could come to this conclusion using Calculus (Lagrange Multipliers), but in this case the A/G means inequality gives a much simpler proof. It\'s easy to see that, for fixed p, A can be made as close to 0 as desired; and, for fixed A, p can be made as big as desired. I\'m almost sure this generalizes to convex polygons with n > 2, at least for those that can be inscribed in a circle, Maybe someone gives this general proof. I don\'t remember now.