Determine among all the triangles with equal perimeter 2p the one with the greatest area. I need to solve the problem but I still haven't found the right way to solve it
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Why $2p$? Why the factor of $2$? How would your problem differ if you had instead simplified and asked about perimeter $p$? And what have you tried? – David G. Stork Jun 01 '23 at 22:56
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Area $T={\sqrt {s(s-a)(s-b)(s-c)}} $ where s=(a+b+c)/2. Using elem. calc. T max when all sides = – herb steinberg Jun 02 '23 at 00:10
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If we are going to use differentiation to get the answer for this it's going to be a real hassle. So instead we will use a neat inequality, which is the AM-GM Inequality.
Let $a,b$ and $c$ be the sides of a triangle with a perimeter of $2p$. By Heron's Formula we have $Area = \sqrt{s(s-a)(s-b)(s-c)}$ where $s$ is the semi-perimeter. If we substitute $p$ as the semi-perimeter we get
$$Area = \sqrt{p(p-a)(p-b)(p-c)}$$
Here $p$ is a constant so what we really want to maximize is the product $(p-a)(p-b)(p-c)$.
Let $x=(p-a), y=(p-b)$ and $z=(p-c)$. The AM-GM inequality states that $\displaystyle\frac{x+y+z}{3}\ge\sqrt[3]{xyz}$. By substitution we get
$$\frac{(p-a)+(p-b)+(p-c)}{3}\ge\sqrt[3]{(p-a)(p-b)(p-c)}$$
$$\implies \frac{3p-(a+b+c)}{3}\ge\sqrt[3]{(p-a)(p-b)(p-c)}$$
$$\implies \frac{p}{3}\ge\sqrt[3]{(p-a)(p-b)(p-c)} \ (\because a+b+c = perimeter = 2p)$$
Thus we can see that if $\displaystyle a=b=c=\frac{2p}{3}$ we get the maximum for the product $(p-a)(p-b)(p-c)$ which is $\displaystyle\frac{8p^3}{27}$. Therefor the maximum area is $\displaystyle \sqrt{\frac{8p^4}{27}}$
Aleph Null
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