Using Lagrange Multipliers to Find Extrema of $f$

Question

I have been trying to solve this lagrange problem for an hour with no result. I can't seem to isolate any variable.

$$f(x,y) = y^2 -4xy +4x^2$$

With constraint:

$$x^2 + y^2 = 1$$

I have found $x$ to = $-\left(\frac{λ-4}{2}\right)^2 + 1$. Using the constraint $x$ is also equal to $\sqrt{1-y^2}$. This still leaves me with two unknowns. I have tried solving for lambda and setting them equal to each other and I end up with $x^2 = y^2-\frac{3xy}{2}$.

There are still two unknown variables here. I am not sure what I should be isolating to get an answer to work with.

Answer 1

I would say the most direct way to do this would be to but this into matrix form.

$f(x,y)= \mathbf x^T A \mathbf x$

That is $f(x,y)= \begin{bmatrix} x&y \end{bmatrix} \begin{bmatrix} 4&-2\\-2&1 \end{bmatrix}\begin{bmatrix} x\\y \end{bmatrix}$

And the eigenvalues:

$\lambda$ is an eigevalue and $\mathbf v$ is and eigenvector if

$A\mathbf v = \lambda \mathbf v\\ (A-\lambda I)\mathbf v = \mathbf 0\\ \det (A-\lambda I) = 0$

$A-\lambda I = \begin{bmatrix} 4-\lambda&-2\\-2&1-\lambda \end{bmatrix}\\ \det (A-\lambda I) = \lambda^2 - 5\lambda = 0$

$\lambda = 0,5$ which will be the min and the max. And the normalized eigenvalues are the coordiniates that maximise and minimize.

$\lambda = 0\\ A\mathbf v = \mathbf 0\\ \begin{bmatrix} 4&-2\\-2&1 \end{bmatrix}\mathbf v = 0\\ \mathbf v = \begin{bmatrix} 1\\2 \end{bmatrix}$

$\lambda = 5\\ (A-2I)\mathbf v = \mathbf 0\\ \begin{bmatrix} -1&-2\\-2&-4 \end{bmatrix}\mathbf v = 0\\ \mathbf v = \begin{bmatrix} -2\\1 \end{bmatrix}$

But that isn't Lagrange multipliers:

Before getting to far, notice:

$f(x,y) = (y-2x)^2$

$\forall x,y, f(x,y) \ge 0$

There is an $(x,y)$ pair that satisfies the constraint such that $f(x,y) = 0$

$(\frac {1}{\sqrt5},\frac {2}{\sqrt5}),(-\frac {1}{\sqrt5},-\frac {2}{\sqrt5})$

Method of Lagrange multipliers:

$F(x,y,\lambda) = (y-2x)^2 + \lambda(x^2+y^2 -1 )\\ \frac {\partial F}{\partial x} = -4(y-2x) + 2\lambda x = 0\\ \frac {\partial F}{\partial y} = 2(y-2x) + 2\lambda y = 0\\ $

Adding twice the second to the first we get

$2\lambda x + 4\lambda y = 0$

Either $\lambda = 0$ and $y = 2x.$ This is the minimum we found above.

$x = -2y$ must be our maximum. Plug it into the constraint ...

$(\frac{2}{\sqrt 5},-\frac {1}{\sqrt 5}),(-\frac{2}{\sqrt 5},\frac {1}{\sqrt 5})$

$f(\frac{2}{\sqrt 5},-\frac {1}{\sqrt 5}) = 5$

Answer 2

You are given $f(x,y)= y^2- 4xy+ 4x^2$ with constraint $g(x, y)= x^2+ y^2= 1$. Then $\nabla f= <-4y+ 8x, 2y- 4x>$ and $\nabla g= <2x, 2y>$. The basic idea of the Lagrange multiplier method is that those two vectors are in the same direction- one is a multiple of the other: $<-4y+ 8x, 2y- 4x>= \lambda<2x, 2y>$. Equating the same components, we have the two equations $4x+ 8y= 2\lambda x$ and $2y- 4x= 2\lambda y$. Together with the constraint $x^2+ y^2= 1$ that gives three equations for the three unknowns x, y, and $\lambda$.

But a value for $\lambda$ is not necessary for this problem so a good way to solve these equations is to first eliminate $\lambda$ by dividing one of the first two equations by the other: $\frac{4x+ 8y}{2y- 4x}= \frac{2x+ 4y}{y- 2x}= \frac{x}{y}$. From that $y(2+ 4y)= x(y- 2x)$ so $2y^2+ 4xy= xy- 2x^2$, $2y^2+ 3xy+ 2x^2= 0$. Completing the square, $2(x^2+ (3/2)xy+ (9/16)y^2-$$(9/16)y^2)+ y^2= 2(x+ (3/4)y)^2- (9/8)y^2= 0$. x+ (3/4)y= (3/4)y$.