# MIT OCW

## 18.02 Multivariable Calculus

### Lagrange Multiplier

We've known the method to maximize or minimize a multivariable function, but what happens if there are some constraint? The critical point usually does not fulfill the constraint, so we have to maximize or minimize in another way.

#### Introduction

**e.g.** Find the closest point to origin on `$xy=3$`

.

**Idea** We are going to minimize `$f(x,y)=x^2+y^2$`

subject to `$xy=3$`

. These are the level surfaces of `$f(x,y)$`

and `$g(x,y)=xy=3$`

.

When the level surface `$f(x,y)=c$`

becomes smaller and smaller, until it has no intersection with the `$xy=3$`

, then we have almost achieved the goal. We can find that at the maximum or minimum `$f_0$`

, the level surface `$f(x,y)=f_0$`

is tangent to the level surface `$g(x,y)=3$`

.

It means that the gradient at `$f(x,y)$`

is parallel to `$g(x,y)$`

, namely, `$\nabla f=\lambda\nabla g$`

, where `$\lambda$`

is an unknown. So, we have such system of equations `$$\left\{\begin{aligned}2x=\lambda y\\2y=\lambda x\\xy=3\end{aligned}\right.,$$`

and we get the point is `$(\sqrt3,\sqrt3)$`

or `$(-\sqrt3,-\sqrt3)$`

.

#### Lagrange Multiplier

Now we conclude the method, given a function `$f(x,y)$`

and a constraint `$g(x,y)=c$`

, then to maximize or minimize the function, we need to solve the system of equations `$\left\{\begin{aligned}\frac{\partial f}{\partial x}&=\lambda\frac{\partial g}{\partial x}\\\frac{\partial f}{\partial y}&=\lambda\frac{\partial g}{\partial y}\\g(x,y)&=c\end{aligned}\right..$`

#### Geometry

Why it is correct? If there is no constraint `$g(x,y)=c$`

, we just solve `$f_x=f_y=0$`

, and it means when we move on a horizontal surface near the point, the function doesn't change.

Now we have a constraint, similarly, we just find a point where we move along the constraint surface that the function doesn't change too. So `$\nabla_{\hat\mathbf{u}}f=0$`

, where `$\hat\mathbf{u}$`

is any direction on the constraint surface, in other words, `$\nabla f\cdot\hat\mathbf{u}=0$`

. Since the gradient `$\nabla g$`

is also normal to the constraint surface, we have `$\nabla f\parallel\nabla g$`

.

#### Application

**e.g.** Find a best solution to minimizing the surface area of a pyramid with a given triangular base `$a_1,a_2,a_3$`

and a given height `$h$`

.

**Solution** We can plot the pyramid in `$xy$`

plane.

And it looks like the following in 3D space.

To determine the position of the vertex `$D$`

more easily, we project the vertex on the `$xy$`

plane.

And take the distance from the projection to three sides `$a_1,a_2,a_3$`

as `$d_1,d_2,d_3$`

, and we can express the surface area `$S$`

and the base area `$A$`

: `$$\begin{aligned}S=\frac12a_1\sqrt{d_1^2+h^2}+\frac12a_2\sqrt{d_2^2+h^2}+\frac12a_3\sqrt{d_3^2+h^2}\\A=\frac12a_1d_1+\frac12a_2d_2+\frac12a_3d_3\end{aligned}.$$`

And apply the Lagrange Multiplier `$$\left\{\begin{aligned}\frac{\partial S}{\partial d_1}=\lambda\frac{\partial A}{\partial d_1}\\\frac{\partial S}{\partial d_2}=\lambda\frac{\partial A}{\partial d_2}\\\frac{\partial S}{\partial d_3}=\lambda\frac{\partial A}{\partial d_3}\end{aligned}\right.,$$`

and found that `$d_1=d_2=d_3$`

, so the vertex is just above the incenter of the triangular base.