Geometric maxima minima — a little question for IITJEE Mains


Of all triangles with a given perimeter, find the one with maximum area.


Consider an arbitrary triangle with side lengths a, b, c and perimeter 2s=a+b+c. By Heron’s formula, its area F is given by

F = \sqrt{s(s-a)(s-b)(s-c)}

Now, the arithmetic mean geometric mean inequality gives

\sqrt[3]{s(s-a)(s-b)(s-c)} \leq \frac{(s-a)+(s-b)+(s-c)}{3}=\frac{s}{3}

Therefore, F \leq \sqrt{s(\frac{s}{3})^{3}}=s^{2}\frac{\sqrt{3}}{9}

where inequality holds if and only if s-a=s-b=s-c, that is, when a=b=c.

Thus, the area of any triangle with perimeter 2s does not exceed \frac{s^{2}\sqrt{3}}{9} and is equal to \frac{s^{2}\sqrt{3}}{9} only for an equilateral triangle. QED.

More later,

Nalin Pithwa

PS: Ref: Geometric Problems on Maxima and Minima by Titu Andreescu et al.


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