11. Lagrange Multipliers

12 - Lagrange Multipliers

Lagrange Multiplier

Let $f(x,y,z)$ and $F(x,y,z)$ be two functions on an open subset $U\subset {\mathbb{R}}^3$
Suppose $f,F$ have continuous partial derivatives and that $\nabla F\ne 0$ on $U$

Let $(x_0,y_0,z_0)\in U$ be a local minimum or maximum of $f(x,y,z)$ where $F(x,y,z)=0$

Then

$$\text{there exists }\lambda \text{ such that }\nabla f(x_0,y_0,z_0)=\lambda \nabla (x_0,y_0,z_0)$$

where $\lambda$ is known as the Lagrange multiplier

Proof

Since $\nabla F\ne 0$ on $U$ then let

$$S=\{(x,y,z)\in U:F(x,y,z)=0\}$$

where $S$ is a surface

From assumption, $(x_0,y_0,z_0)\in S$ and also a local minimum or maximum

For any curve $\mathbf{v}(t)=(x(t),y(t),z(t))$ on surface $S$ and passing $(x_0,y_0,z_0)$

$$F(\mathbf{v}(t))=0\forall t\in (-ϵ,ϵ)\text{ with }\mathbf{v}(0)=(x_0,y_0,z_0)$$

Let $h(t)=f(\mathbf{v}(t))$
By definition, $t=0$ is a local minimum/maximum of $h(t)$ hence $h^{\prime }(t)=0$

Hence

$$h^{\prime }(0)=\nabla f(\mathbf{v}(0))\cdot {\mathbf{v}}^{\prime }(0)=0$$

So $\nabla f(x_0,y_0,z_0)$ is perpendicular to ${\mathbf{v}}^{\prime }(0)$

Since $\mathbf{v}(t)$ is a curve lying on surface $S$ then ${\mathbf{v}}^{\prime }(0)$ is any tangent vector to $S$ at $(x_0,y_0,z_0)$
Therefore $\nabla f(x_0,y_0,z_0)=0$ or $\nabla f(x_0,y_0,z_0)\ne 0$ and is normal to $S$ at $(x_0,y_0,z_0)$

As we know $\nabla F(x_0,y_0,z_0)$ is normal to $S$ at $(x_0,y_0,z_0)$ then

$$\nabla f(x_0,y_0,z_0)\text{ and }\nabla F(x_0,y_0,z_0)\text{ are parallel}$$

Since $\nabla F(x_0,y_0,z_0)\ne 0$ then there exists $\lambda$ such that

$$\nabla f(x_0,y_0,z_0)=\lambda \nabla F(x_0,y_0,z_0)$$

Solving for Lagrange Multipliers

Let

$$G(x,y,z,\lambda )=f(x,y,z)-\lambda F(x,y,z)$$

Then we need to find when

$$\frac{\partial G}{\partial x}=\frac{\partial G}{\partial y}=\frac{\partial G}{\partial z}=\frac{\partial G}{\partial \lambda }=0$$

So a solution to $(x,y,z,\lambda )$ is a critical point of $G(x,y,z,\lambda )$

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