If perimeter of a triangle is $2d$, what is the length of sides so the triangle has maximal area?

I found some solution using circle and angles, but I think I have to use derivatives.

I need help.

  • 1
    $\begingroup$ If you want to use derivatives to solve it, try Heron's formula. It will be with two indeterminates, so going down that road might be difficult if you do not know how to do work with gradients. I suspect, however, that the equilateral triangle is the answer. $\endgroup$
    – Arthur
    Commented Jun 8, 2013 at 12:53
  • $\begingroup$ I tried with that formula, but I have problems to solve it. $T=\sqrt {s(s-a)(s-b)(s-c)}$ and $s=\frac {a+b+c}{2}=\frac {2d}{2}=d$ and $c=2d-a-b$, I got that $T=\sqrt {d(d-a)(d-b)(a+b-d)}$. At the end, I got that $T^2=-d^4+(a+b)2d^3-(a^2+3ab+b^2)d^2+(a^2b+ab^2)d$. $\frac {\partial T^2}{\partial a}=2d^3-2d^2a-3bd^2+2abd+b^2d=0$ $\frac {\partial T^2}{\partial b}=2d^3-3ad^2-2bd^2+a^2d+2abd=0$. If I multiply one of those equations and add to another, $2d^3$ and $2abd$ will cancel. But I don't know how to finish this and calculate $a$ and $b$. $\endgroup$
    – user23709
    Commented Jun 8, 2013 at 13:24
  • $\begingroup$ If you set your expression for $\frac {\partial T^2}{\partial a}$ equal to zero and solve for $a$ you will get $a = \frac{2d-b}{2}$. You can then substitute that for $a$ in your original $T^2$ expression, differentiate with respect to $b$ and set that to zero. $\endgroup$ Commented Jun 8, 2013 at 17:24
  • $\begingroup$ I did it and I got $a=c=\frac{5d}{6}$, $b=\frac{d}{3}$ $\endgroup$
    – user23709
    Commented Jun 8, 2013 at 20:16
  • $\begingroup$ See also Maximal area of a triangle and Proving the regular n-gon maximizes area for fixed perimeter. $\endgroup$
    – AakashM
    Commented Dec 19, 2013 at 13:40

3 Answers 3


Let $a$, $b$ and $c$ be the sides of a triangle. The perimeter, $p=a+b+c$, is fixed and we want to find the values of $a$, $b$ and $c$ that give the triangle maximum area. Heron's formula says that the triangle's area is $$A=\sqrt{s(s-a)(s-b)(a-c)}$$ where s is the semiperimeter $\frac{a+b+c}{2}=\frac{p}{2}.$

Because p is fixed, we can write $c=p-a-b$. Substituting this into the equation above and squaring we find that

\begin{eqnarray} 16A^2=p(p-2a)(p-2b)(2a+2b-p). \quad\quad(1) \end{eqnarray}

In the first part, we fix $a$ and see what we can do with $b$ to get a maximum. To this end, differentiating with respect to $b$ gives \begin{align*} \nonumber 32A\frac{dA}{db}=p(p-2a)\left[(p-2b)(2)+(2a+2b-p)(-2)\right] \nonumber =4p(p-2a)(p-2b-a). \end{align*} If we set this equal to $0$ to find the critical points we find there are two possibilities. The first is that $p=2a$ which leads to $a=\frac{p}{2}$ and $b=c=\frac{p}{4},$ which do not make a proper triangle.
The more interesting possibility is that $p-2b-a=0$, or that $b=\frac{p-a}{2}.$ The significance of this value for $b$ becomes apparent when we see that $c=p-a-b=p-a-\frac{p-a}{2}=\frac{p-a}{2}=b.$ Thus we have established that the triangle is at least iscosceles.

In the second part, we use the value of $b$ just obtained, and see what we can do with $a$. Substituting for $b$ in (1) we find that $$16A^2=p(p-2a)a^2$$ which we differentiate with respect to $a$ to get $$32A\frac{dA}{da}=p\left[(p-2a)(2a)+a^2(-2)\right] = 2ap(p-3a).$$ Setting this equal to 0 again gives us $a=\frac{p}{3}.$ Substituting back, we find that $b = \frac{p-a}{2}=\frac{p}{3}$ and finally $c=p-a-b=\frac{p}{3}$ as well, proving that maximum area is achieved when the triangle is equilateral.

  • $\begingroup$ why $a=\frac{p}{2}$ and $b=c=\frac{p}{4}$ doesn't make proper triangle? $\endgroup$
    – user23709
    Commented Jun 8, 2013 at 20:47
  • 1
    $\begingroup$ Because $a = b+c$, contravening the Triangle inequality. $\endgroup$ Commented Jun 8, 2013 at 22:56
  • $\begingroup$ If we said, after "Thus we have established that the triangle is at least iscosceles. ", simply: "By symmetry [since we could just run the same argument with $a = p - b - c$, etc], the triangle is therefore equilateral", would that be too hand-wavey? $\endgroup$
    – AakashM
    Commented Dec 19, 2013 at 13:41
  • $\begingroup$ @PeterPhipps In the last sentence, showing that c=a is unnecessary since that is covered transitively by b=a through b=c in the former paragraph. $\endgroup$
    – Amos C N
    Commented Nov 3, 2022 at 3:50

Here is another way. Suppose that the lengths of the sides of the triangle are $a$,$b$ and $c$ such that that the perimeter of the triangle is fixed and it is $2s$. Using the Heron's formula, the area, $A$ is $$A=\sqrt{s(s-a)(s-b)(s-c)}$$ The AM-GM inequality for three positive reals $a,b,c$ states that $\displaystyle (abc)^\frac{1}{3}\leq \frac{a+b+c}{3} $wih equality at $a=b=c$.We may use it above on $A$.


Area is $\sqrt{s (s-a) (s-b) (s-c)}$

if $s$ is $(a+b+c)/2$

since we are trying to maximize the Area, we could say we are trying to maximize the area squared so just maximize

$$s (s-a) (s-b) (s-c)$$

since s is a constant because the perimeter is a fixed quantity, this is equivalent to maximizing $(s-a) (s-b) (s-c)$

but also note $(s-a) + (s-b) + (s-c) = 3s - \text{perimeter} = s$

which is a constant

so, by AM-GM where $xyz$ is the quantity, we are maximizing and $x+y+z$ is a constant $x=y=z$

so, $(s-a)=(s-b)=(s-c)$

so, $a=b=c$

so, the triangle is equilateral.

  • $\begingroup$ Nice variant with no derivative. $\endgroup$
    – mins
    Commented Oct 24, 2022 at 14:59

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