Q.A technical company is making rectangular solar panels. It has 300 meters of material to fence the perimeter for installing them on the roof. The design of the panel has a division parallel to one side which divides the panel into two parts. Let the length of the side perpendicular to the division be $x$ meters and the length of the side parallel to it be $y$ meters.

Lagrange Multipliers: Optimising Under a Constraint

To find the highest point on a hill you just walk uphill until every direction goes down — that is unconstrained optimisation. Now suppose you must stay on a footpath through the park. The highest point on the hill may lie off the path, so you want the highest point that lies on the path. That is constrained optimisation, and Lagrange multipliers are the tool for it: maximise or minimise $f(x,y)$ subject to a constraint $g(x,y)=0$.

The Idea: Two Gradients Line Up

At the best point on the constraint curve, you cannot move along the curve and still increase $f$ — otherwise you would keep going. So the direction of steepest ascent, $\nabla f$, has no component along the curve; it points straight across it. The curve's own normal direction is $\nabla g$. Hence at the optimum the two gradients are parallel:

The scalar $\lambda$ is the Lagrange multiplier. It measures how much the optimal $f$ would change if the constraint were loosened by one unit.

The Method

To find the extrema of $f(x,y)$ subject to $g(x,y)=0$, form the Lagrangian

and solve the system

The third equation just restates $g(x,y)=0$; the first two give $\nabla f = \lambda\nabla g$. Every solution is a candidate — evaluate $f$ at each to decide which is the maximum and which the minimum.

A Worked Example

Find the point on the circle $x^2+y^2=1$ that maximises $f(x,y)=x+y$.

Here $g = x^2+y^2-1$, so $\nabla f=(1,1)$ and $\nabla g=(2x,2y)$. The equations $1=2\lambda x$ and $1=2\lambda y$ give $x=y$. The constraint then forces $2x^2=1$, so $x=y=\pm\tfrac{1}{\sqrt2}$.

At $\left(\tfrac{1}{\sqrt2},\tfrac{1}{\sqrt2}\right)$, $f=\sqrt2$ (the maximum); at $\left(-\tfrac{1}{\sqrt2},-\tfrac{1}{\sqrt2}\right)$, $f=-\sqrt2$ (the minimum).

Common Mistakes …