# Finding the triangle with the maximum area with a given perimeter

Question: Of all triangles with a given perimeter, find the triangle with the maximum area. Justify your answer.

My approach: Let us have any $$\Delta ABC$$ such that the sides opposite to $$A$$ is of length $$a$$, the side opposite to $$B$$ is of length b and the side opposite to $$C$$ is of length $$c$$. Now since the perimeter to $$\Delta ABC$$ is fixed, thus we must have $$P=a+b+c$$ to be constant.

Now fix any side of the triangle, let us fix $$BC$$. Therefore $$b+c=P-a$$, which implies that $$b+c$$ is constant. Thus the locus of the point $$A$$ must be an ellipse having one of it's axis as the side $$BC$$. Now let us select any point $$A$$ on the ellipse and drop a perpendicular to the axis $$BC$$. Let it meet the major axis at $$P$$. Now let $$AP=h$$. Therefore, the area of the $$\Delta ABC=\frac{1}{2}.h.BC=\frac{1}{2}.h.a.$$ Now since $$\frac{1}{2}a$$ is constant, implies the area of $$\Delta ABC$$ can be maximized by maximizing $$h$$. Now clearly $$h$$ attains it's maximum value when it coincides with the other axis of the ellipse under consideration, that is when $$AB=AC$$. Thus $$\Delta ABC$$ must be isosceles with $$AB=AC$$ to get a maximum value of the area of $$\Delta ABC$$.

Thus it is clear that the triangle which will have the maximum area must be one of the isosceles triangles $$ABC$$ having $$BC$$ as the base.

Thus in any such $$\Delta ABC$$, we must have $$a+b+c=a+2b \hspace{0.2 cm}(\because b=c)=P\implies b=\frac{1}{2}(P-a).$$ Thus by Heron's formula we have $$|\Delta ABC|=\sqrt{\frac{P}{2}\left(\frac{P}{2}-a\right)\left(\frac{a}{2}\right)\left(\frac{a}{2}\right)}=\frac{a\sqrt{P}}{4}\sqrt{P-2a}.$$

Now to obtain the condition for maximizing $$|\Delta ABC|$$ we have to check when $$\frac{d}{da}\frac{a\sqrt{P}}{4}\sqrt{P-2a}=0.$$ Now $$\frac{d}{da}\frac{a\sqrt{P}}{4}\sqrt{P-2a}=0\\\iff (P-2a)^{1/2}-a(P-2a)^{-1/2}=0\\\iff P-2a=a\\\iff 3a=P\\\iff a=\frac{P}{3}.$$

Now note that $$\frac{d^2}{da^2}|\Delta ABC|<0$$, which implies that $$|\Delta ABC|$$ attain it's maximum value when $$a=\frac{P}{3}.$$ This implies that $$b=c=\frac{P}{3}$$. Thus we have $$a=b=c=\frac{P}{3}$$. Therefore, $$|\Delta ABC|$$ is maximized if and only if $$a=b=c$$, i.e, the triangle is equilateral.

Can someone check if this solution is correct or not? And other solutions are welcomed. Please ensure that the solutions are based on geometry and one-variable calculus. This problem can be solved using Lagrange multipliers or multi-variable calculus, but I do not want a solution using the same.

• Clearly it must be an equilateral triangle, with sides $P/3$. – David G. Stork Apr 5 '20 at 18:47

The Heron's formula for the triangle is

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

where $$s=\frac p2$$. Then, Apply the AM-GM inequality to get

$$A =\sqrt{s(s-a)(s-b)(s-c)} \le \left[ s \left( \frac{3s-(a+b+c)}{3} \right) ^3\right]^{1/2} = \frac{s^2}{3\sqrt3}=\frac{p^2}{12\sqrt3}$$

where the equality, or the maximum area, occurs at $$a=b=c=\frac p3$$.

Just for giving another way to deduce a solution. Take a closed string and leaving a fixed side, say $$a$$, of the triangle draw an ellipse as usual. It is evident that the triangle with the largest area occurs with an isosceles triangle because it has the same base as all but has a greater height (actually a vertical semiaxis of the ellipse).

Now the area of this isosceles triangle is $$A=\dfrac a4\sqrt{(2b)^2-a^2}$$ but because of $$a+2b=p$$ we have a fonction of $$a$$, call it $$x$$, defined by $$A=\frac x4\sqrt{p^2-2px}$$ The derivative of A being equal to $$A'=\frac{p^2-3px}{4\sqrt{p^2-2px}}$$ we see that the maximum area is taken when $$a=\dfrac p3$$ then $$b=\dfrac p3$$ too.

Area of triangle with vertices $$a,b,c$$ $$\frac12(a\times b+b\times c+c\times a)=\frac12\left(a^R\cdot b+b^R\cdot c+c^R\cdot a\right)\tag1$$ where $$(x,y)^R=(-y,x)$$ is $$\frac\pi2$$ rotation counter-clockwise.

Perimeter of triangle $$|a-b|+|b-c|+|c-a|\tag2$$ maximize $$(1)$$ $$a^R\cdot(\delta b-\delta c)+b^R\cdot(\delta c-\delta a)+c^R\cdot(\delta a-\delta b)=0\tag3$$ for all variations that keep $$(2)$$ fixed $$\frac{a-b}{|a-b|}\cdot(\delta a-\delta b)+\frac{b-c}{|b-c|}\cdot(\delta b-\delta c)+\frac{c-a}{|c-a|}\cdot(\delta c-\delta a)=0\tag4$$ we can use $$\delta a-\delta b$$, $$\delta b-\delta c$$, and $$\delta c-\delta a$$ as the variations as long as we take into account their dependence: $$(\delta a-\delta b)+(\delta b-\delta c)+(\delta c-\delta a)=0\tag5$$ Orthogonality requires a point $$\mu$$ and constant $$\lambda$$ so that \begin{align} a^R&=\lambda\frac{b-c}{|b-c|}+\mu^R\\ b^R&=\lambda\frac{c-a}{|c-a|}+\mu^R\\ c^R&=\lambda\frac{a-b}{|a-b|}+\mu^R \end{align}\tag7 That is, $$|a-\mu|=|b-\mu|=|c-\mu|=\lambda\tag8$$ and $$(a-\mu)\cdot(b-c)=(b-\mu)\cdot(c-a)=(c-\mu)\cdot(a-b)=0\tag9$$ Therefore, \begin{align} |a-b|^2 &=|(a-\mu)-(b-\mu)|^2\\ &=|a-\mu|^2+|b-\mu|^2-2(a-\mu)\cdot(b-\mu)\\ &=|a-\mu|^2+|c-\mu|^2-2(a-\mu)\cdot(c-\mu)\\ &=|(a-\mu)-(c-\mu)|^2\\ &=|a-c|^2 \end{align}\tag{10} The step where $$b$$ changes to $$c$$ follows from $$|c-\mu|=|b-\mu|$$ and $$(a-\mu)\cdot(c-b)=0$$.

Similarly $$|a-b|^2=|c-b|^2$$. Thus, the triangle is equilateral.