Mathematical Tools for Physics

8—Multivariable Calculus 230

To implement this picture so that you can compute with it, look at the gradient off and the gradient of
φ. The gradient vector is perpendicular to the curvef=constant orφ=constant. At the point where the curves
are tangent to each other these gradients are in the same direction (or opposite, no matter). One vector is a
scalar times the other.
∇f=λ∇φ

In the second picture, the arrows are the gradient vectors forfand forφ. Break this into components and you
have
∂f
∂x

−λ

∂φ

∂x

= 0,

∂f

∂y

−λ

∂φ

∂y

= 0, φ(x,y) = 0

These are three equations in three unknowns(x,y,λ). These are the equations to solve for the position of the
maximum or minimum value off. You’re looking forxandy, so you’ll be tempted to ignore the third variable
λand to eliminate it. Look again; this parameter, the Lagrange multiplier, has a habit of being significant.

Examples of Lagrange Multipliers
The first example that I mentioned: What is the largest rectangle that you can inscribe in an ellipse? Let the
ellipse and the rectangle be centered at the origin. The upper right corner of the rectangle is at(x,y), then the
area of the rectangle is

Area =f(x,y) = 4xy,

with constraintφ(x,y) =

x^2

a^2

+

y^2

b^2

−1 = 0

The equations to solve are now

∇(f−λφ) = 0, and φ= 0, which become

4 y−λ

2 x

a^2

= 0, 4 x−λ

2 y

b^2

= 0,

x^2

a^2

+

y^2

b^2

−1 = 0 (22)

The solutions to these three equations are straight-forward. They arex=a/

√

2 ,y=b/

√

2 ,λ= 2ab. The
maximum area is then 4 xy= 2ab. The Lagrange multiplier turns out to be the required area. Does this reduce
to the correct result for a circle?