Tag: Lagrange multipliers

Suppose you have a fixed length of wire and want to bend it into a triangle. Which triangle encloses the largest possible area?

It is natural to guess that the answer is the equilateral triangle. But why? This is a beautiful example of how a geometric optimization problem can be turned into a multivariable calculus problem and solved using Lagrange multipliers.

Setting up the problem

Let the side lengths of the triangle be

$$a,b,c.$$

Suppose the perimeter is fixed and equal to $P$. Thus,

$$a+b+c=P.$$

We want to determine which values of $a$, $b$, and $c$ produce the largest possible area.

Heron’s formula

Let

$$s=\frac{P}{2}$$

be the semiperimeter. Heron’s formula can be written in squared form as

$$A^2=s(s-a)(s-b)(s-c).$$

Because the perimeter is fixed, $s$ is also fixed. Moreover, maximizing $A$ is equivalent to maximizing $A^2$. So this form of Heron’s formula is particularly convenient for our problem.

A useful change of variables

Introduce three new variables:

$$x=s-a,y=s-b,z=s-c.$$

The triangle inequalities imply that $x$, $y$, and $z$ are positive.

Adding the three equations gives

$$x+y+z=3s-(a+b+c).$$

Since

$$a+b+c=2s,$$

we obtain the simple constraint

$$x+y+z=s.$$

Heron’s formula now becomes

$$A^2=sxyz.$$

Since $s$ is fixed, maximizing the area is equivalent to maximizing

$$f(x,y,z)=xyz$$

subject to

$$x+y+z=s.$$

The original geometry problem has therefore become a simple question: among three positive numbers with a fixed sum, when is their product largest?

Using Lagrange multipliers

Define

$$f(x,y,z)=xyz$$

and let the constraint function be

$$g(x,y,z)=x+y+z.$$

At a constrained maximum, the gradients of $f$ and $g$ must be parallel:

$$\nabla f=\lambda \nabla g.$$

We have

$$\nabla f=⟨yz,xz,xy⟩$$

and

$$\nabla g=⟨1,1,1⟩.$$

Therefore, the Lagrange multiplier equations are

$$yz=\lambda ,xz=\lambda ,xy=\lambda .$$

Thus,

$$yz=xz=xy.$$

Since $x$, $y$, and $z$ are positive, these equations imply

$$x=y=z.$$

Their sum is $s$, so

$$x=y=z=\frac{s}{3}.$$

Returning to the triangle

Recall that

$$x=s-a,y=s-b,z=s-c.$$

Since $x=y=z$, we obtain

$$a=b=c.$$

Because the perimeter is $P$, each side must therefore have length

$$a=b=c=\frac{P}{3}.$$

Therefore, the triangle of maximum area is the equilateral triangle.

What is the maximum area?

For an equilateral triangle with side length $\frac{P}{3}$, the area is

$$A_{\text{max}}=\frac{\sqrt{3}}{4}{\left(\frac{P}{3}\right)}^2.$$

Therefore,

$$A_{\text{max}}=\frac{\sqrt{3}{P}^2}{36}.$$

Why this argument is interesting

We started with a geometric question about triangles. Heron’s formula converted the area problem into an algebraic one. A simple change of variables then transformed it into the problem of maximizing the product of three positive numbers whose sum is fixed.

Lagrange multipliers reveal the symmetry automatically: at the maximum, the three variables must be equal. Translating that condition back into geometry tells us that the three sides of the triangle must also be equal.

This is one of the appealing features of multivariable calculus: a geometric statement that seems intuitively obvious emerges naturally from an optimization calculation.

Conclusion: Among all triangles with a fixed perimeter, the equilateral triangle has the largest area.

For another surprising connection between equilateral-triangle geometry and a classical geometric object, see The Steiner Inellipse and a Surprising Area Characterization .

The equilateral triangle is distinguished by its symmetry. In three dimensions, the regular tetrahedron has similar geometric elegance. Explore its face areas, volume, and a three-dimensional Pythagorean theorem in The Geometry of a Tetrahedron .