# Lagrange Multipliers

1. Jun 20, 2012

### magorium

1. The problem statement, all variables and given/known data

Show that , the maximum value of function f(x,y) = x^2 + y^2 is 70 and minimum value is 20
in constraint below.

2. Relevant equations

Constraint : 3x^2 + 4xy + 6y^2 = 140

3. The attempt at a solution

Book's solution simply states the Lagrange rule as :
h(x , y ; L) = x^2 + y^2 - L(3x^2 + 4xy + 6y^2 - 140)

Takes partial derivatives for x , y and L.

h's partial derivative for x = 2x + L(6x + 4y) = 0
h's partial derivative for y = 2y + L(4x + 12y) = 0
h's partial derivative for L = 3x^2 + 4xy + 6y^2 -140 = 0

Then he takes the coefficients determinant of first two equations ( h derv for x , h derv for y)

|1+3L 2L |
|2L 1+6L |

And makes this determinant equal to zero , finds values for L.

The part i don't understand is , why he uses Cramer rule to the first two equations and equals it to zero ?

2. Jun 20, 2012

### algebrat

It is not clear to me that the book is using Cramer's rule, can you show more steps if it is indeed using Cramer's rule.

If they are not using Cramer's rule, then this is how I would interpret what is happening. From the first two equations, book want soutions to A[x y]=0, where [x y] is meant to be a column matrix, and A is a matrix depending on the value L.

We have the trivial solution x=y=0. To find a nontrivial solution, we note that would imply that det(A)=0. That gives us a polynomial in L to help us find candidate values for L, and continue solving equations with those candidates.

EDIT: I just looked up Cramer's rule, it looks like you need determinant of A nonzero (else the denominator is zero), which is not something we need.

3. Jun 20, 2012

### Ray Vickson

Can x and y both = 0? If you agree they cannot both = 0 then you need to have a nonzero solution (x,y) of the linear equations
$$(6L+2) x + 4L y = 0\\ 4L x + (12L+1) y = 0$$
What conditions do you need to satisfy in order that the solution is not x=y=0?

RGV

4. Jun 20, 2012

### algebrat

In fact, I do not completely understand what you are asking, please clarify.

5. Jun 20, 2012

### magorium

algebrat you are right about that. Thanks for it. det(A)=0 actually comes from Cramer Rule for me. Since for finding the first unknown we use the special determinant for it divided by det(A) so since det(A) is 0 , that creates a uncertainty which avoids us finding a trivial solution. That's how i remember the det(A)=0 thing so that's possibly why i called it Cramer Rule :) Thanks for it.