Monthly Archives: November 2019

Deriving Lagrange's multiplier

We wish to consider a special type of optimization problem:

Find the maximum (or minimum) of a function f(x,y) subject to the condition g(x,y)=0\quad\quad(1)

If it is possible to solve g(x)=0 for y so that it is expressed explicitly as y=\psi(x), by substituting y in (1), it becomes

Find the maximum (or minimum) of a single variable function f(x, \psi(x)).

In the case that y can not be obtained from solving g(x,y)=0, we re-state the problem as:

Find the maximum (or minimum) of a single variable function z=f(x,y) where y is a function of x, implicitly defined by g(x, y)=0\quad\quad\quad(2)

Following the traditional procedure of finding the maximum (or minimum) of a single variable function, we differentiate z with respect to x:

\frac{dz}{dx} = f_x(x,y) + f_y(x,y)\cdot \frac{dy}{dx}\quad\quad\quad(3)

Similarly,

g_x(x,y) + g_y(x,y)\cdot \frac{dy}{dx}=0\quad\quad\quad(4)

By grouping g(x, y)=0 and (3), we have

\begin{cases} \frac{dz}{dx}= f_x(x, y)+g_x(x, y)\cdot \frac{dy}{dx}\\ g(x,y) = 0\end{cases}\quad\quad\quad(5)

The fact that \frac{dz}{dx}= 0 at any stationary point x^* means for all (x^*, y^*) where g(x^*, y^*)=0,

\begin{cases} f_x(x^*, y^*)+g_x(x^*, y^*)\cdot \frac{dy}{dx}\vert_{x=x^*}=0 \\ g(x^*,y^*) = 0\end{cases}\quad\quad\quad(6)

If g_y(x^*,y^*) \ne 0 then from (4),

\frac{dy}{dx}\vert_{x=x^*} = \frac{-g_x(x^*, y^*)}{g_y(x^*, y^*)}

Substitute it into (6),

\begin{cases} f_x(x^*, y^*)+f_y(x^*, y^*)\cdot (\frac{-g_x(x^*, y^*)}{g_y(x^*, y^*)})=f_x(x^*, y^*)+g_x(x^*, y^*)\cdot (\frac{-f_y(x^*, y^*)}{g_y(x^*, y^*)})\\ g(x^*,y^*) = 0\end{cases}\quad\quad\quad(7)

Let \lambda = \frac{-f_y(x^*, y^*)}{g_y(x^*, y^*)}, we have

f_y(x^*, y^*) + \lambda g_y(x^*, y^*) =0\quad\quad\quad(8)

Combining (7) and (8) gives

\begin{cases} f_x(x^*, y^*)+\lambda g_x(x^*, y^*) = 0 \\ f_y(x^*, y^*)+\lambda g_y(x^*, y^*)=0 \\ g(x^*, y^*) = 0\end{cases}

It follows that to find the stionary points of z, we solve

\begin{cases} f_x(x, y)+\lambda g_x(x, y) = 0 \\ f_y(x, y)+\lambda g_y(x, y)=0 \\ g(x, y) = 0\end{cases}\quad\quad\quad(9)

for x, y and \lambda.

This is known as the method of Lagrange’s multiplier.

Let F(x,y,\lambda) = f(x,y) + \lambda g(x,y).

Since

F_x(x,y,\lambda) = f_x(x,y) + \lambda g_x(x,y),

F_y(x,y,\lambda)=f_y(x,y) + \lambda  g_y(x,y),

F_{\lambda}(x,y,\lambda) = g(x, y),

(9) is equivalent to

\begin{cases} F_x(x, y, \lambda)=0 \\ F_y(x,y,\lambda)=0 \\ F_{\lambda}(x, y) = 0\end{cases}\quad\quad\quad(10)

Let’s look at some examples.

Example-1 Find the minimum of f(x, y) = x^2+y^2 subject to the condition that x+y=4

Let F(x, y, \lambda) = x^2+y^2+\lambda(x+y-4).

Solving

\begin{cases}F_x=2x-\lambda=0  \\ F_y = 2y-\lambda = 0 \\ F_{\lambda} = x+y-4=0\end{cases}

for x, y, \lambda gives x=y=2, \lambda=4.

When x=2, y=2, x^2+y^2=2^2+2^2=8.

\forall  (x, y) \ne (2, 2), x+y=4, we have

(x-2)^2 + (y-2)^2 > 0.

That is,

x^2-4x+4 + y^2-4y+4 = x^2+y^2-4(x+y)+8 \overset{x+y=4}{=} x^2+y^2-16+8>0.

Hence,

x^2+y^2>8, (x,y) \ne (2,2).

The target function x^2+y^2 with constraint x+y=4 indeed attains its global minimum at (x, y) = (2, 2).

I first encountered this problem during junior high school and solved it:

(x-y)^2 \ge 0 \implies x^2+y^2 \ge 2xy

x+y=4\implies (x+y)^2=16\implies x^2+2xy +y^2=16\implies 2xy=16-(x^2+y^2)

x^2+y^2 \ge 16-(x^2+y^2) \implies x^2+y^2 \ge 8\implies z_{min} = 8.

I solved it again in high school when quadratic equation is discussed:

x+y=4 \implies y =4-x

z=x^2+y^2 \implies z = x^2+(4-x)^2 \implies 2x^2-8x+16-z=0

\Delta =  64-4 \cdot 2\cdot (16-z) \ge 0 \implies z \ge 8\implies z_{min} = 8

In my freshman calculus class, I solved it yet again:

x+y=4 \implies y=4-x

z = x^2+(4-x)^2

\frac{dz}{dx} = 2x+2(4-x)(-1)=2x-8+2x=4x-8

\frac{dz}{dx} =0  \implies x=2

\frac{d^2 z}{dx^2} = 4 > 0 \implies x=2, z_{min}=2^2+(4-2)^2=8

Example-2 Find the shortest distance from the point (1,0) to the parabola y^2=4x.

We minimize f = (x-1)^2+y^2 where y^2=4x.

If we eliminate y^2 in f, then f = (x-1)^2+4x. Solving \frac{df}{dx} = 2x+2=0 gives x=-1, Clearly, this is not valid for it would suggest that y^2=-4 from y^2=4x, an absurdity.

By Lagrange’s method, we solve

\begin{cases} 2(x-1)-4\lambda=0 \\2y\lambda+2y = 0 \\y^2-4x=0\end{cases}

Fig. 1

The only valid solution is x=0, y=0, k=-\frac{1}{2}. At (x, y) = (0, 0), f=(0-1)^2+0^2=1. It is the global minimum:

\forall (x, y) \ne (0, 0), y^2=4x \implies x>0.

(x-1)^2+y^2 \overset{y^2=4x}{=}(x-1)^2+4x=x^2-2x+1+4x=x^2+2x+1\overset{x>0}{>}1=f(0,0)

Example-3 Find the shortest distance from the point (a, b) to the line Ax+By+C=0.

We want find a point (x_0, y_0) on the line Ax+By+C=0 so that the distance between (a, b) and (x_0, y_0) is minimal.

To this end, we minimize (x_0-a)^2+(y_0-b)^2 where Ax_0+By_0+C=0 (see Fig. 2)

Fig. 2

We found that

x_0=\frac{aB^2-bAB-AC}{A^2+B^2}, y_0=\frac{bA^2-aAB-BC}{A^2+B^2}

and the distance between (a, b) and (x_0, y_0) is

\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}\quad\quad\quad(11)

To show that (11) is the minimal distance, \forall (x, y) \ne (x_0, y_0), Ax+By+C=0.

Let d_1 = x-x_0, d_2=y-y_0, we have

x = x_0 + d_1, y=y_0 + d_2, d_1 \ne 0, d_2 \ne 0.

Since Ax+By+C=0,

A(x_0+d_1)+B(y_0+d_2)+C=Ax_0+Ad_1+By_0+Bd_2+C=0

That is

Ax_0+By_0+C+Ad_1+Bd_2=0.

By the fact that Ax_0+By_0+C=0, we have

Ad_1 + Bd_2 =0\quad\quad\quad(12)

Compute (x-a)^2+(y-b)^2 - ((x_0-a)^2+(y_0-b)^2) (see Fig. 3)

Fig. 3

yields

\boxed{-\frac{2d_2BC}{B^2+A^2}-\frac{2d_1AC}{B^2+A^2}}+[\frac{d_2^2B^2}{B^2+A^2}]-\underline{\frac{2bd_2B^2}{B^2+A^2}}+(\frac{d_1^2B^2}{B^2+A^2})-\frac{2ad_2AB}{B^2+A^2}-\underline{\frac{2bd_1AB}{B^2+A^2}}+[\frac{d_2A^2}{B^2+A^2}]+(\frac{d_1^2A^2}{B^2+A^2})-\frac{2ad_1A^2}{B^2+A^2}

After some rearrangement and factoring, it becomes

\frac{-2C}{A^2+B^2}(Ad_1+Bd_2)+\frac{-2B}{A^2+B^2}(Ad_1+Bd_2)+\frac{-2A}{A^2+b^2}(Ad_1+Bd_2) + d_1^2+d_2^2

By (12), it reduces to

d_1^2 + d_2^2.

This is clearly a positive quantity. Therefore,

\forall (x, y) \ne (x_0, y_0), Ax+By+C=0 \implies (x-a)^2+(y-b)^2> (x_0-1)^2+(y_0-b)^2