Constrained extreme problem

**Ulysses** · November 5th 2010, 12:34 PM

I have some trouble with this exercise. It says as follows: Find the constrained extremes for $z=2x^2+y^2$ over $\cos (x^2+y^2)-1=0$

I think that difficult of this problem is that I have a family of circles that satisfies $\cos (x^2+y^2)-1=0$ , the family of circles with radius $r=\sqrt[ ]{k \pi}$ .

This is what I did.

$\cos (x^2+y^2)-1=0\Rightarrow{x^2+y^2=k \pi},k\in{Z}$
$F(x,y,\lambda)=2x^2+y^2-\lambda(cos(x^2+y^2)-1)=0$

$\begin{Bmatrix}F_x=4x+2\lambda x \sin(x^2+y^2)=0 & & (1)\\F_y=2y+2\lambda y \sin(x^2+y^2)=0 & & (2)\\F_{\lambda}=\cos(x^2+y^2)-1=0 & & (3)\end{matrix}$

From (1) and (2): $4x-2y+(2\lambda \sin(x^2+y^2))(x-y)=0\Longrightarrow{2\lambda\sin(x^2+y^2)=\display style\frac{4x-2y}{x-y}}$ (*)

Replacing with (*) in (1)

$4x+x \displaystyle\frac{4x-2y}{x-y}}=0\Rightarrow{4x(x-y)+x(4x-2y)=0}\Longrightarrow{x^2-4xy=0\longrightarrow{y=\displaystyle\frac{-x}{4}}}$

Replacing in (3)

$x^2+(\displaystyle\frac{-x}{4})^2=k\pi\Rightarrow{x=\pm{\sqrt[ ]{\displaystyle\frac{16}{17}k\pi}}}$

Well, I'm not sure if this is okey. If it is, how should I proceed?
Thanks for posting!

**Opalg** · November 5th 2010, 01:59 PM

Originally Posted by Ulysses

I have some trouble with this exercise. It says as follows: Find the constrained extremes for $z=2x^2+y^2$ over $\cos (x^2+y^2)-1=0$

I think that difficult of this problem is that I have a family of circles that satisfies $\cos (x^2+y^2)-1=0$ , the family of circles with radius $r=\sqrt[ ]{k \pi}$ .

This is what I did.

$\cos (x^2+y^2)-1=0\Rightarrow{x^2+y^2=k \pi},k\in{Z}$
$F(x,y,\lambda)=2x^2+y^2-\lambda(cos(x^2+y^2)-1)=0$

$\begin{Bmatrix}F_x=4x+2\lambda x \sin(x^2+y^2)=0 & & (1)\\F_y=2y+2\lambda y \sin(x^2+y^2)=0 & & (2)\\F_{\lambda}=\cos(x^2+y^2)-1=0 & & (3)\end{matrix}$

From (1) and (2): $4x-2y+(2\lambda \sin(x^2+y^2))(x-y)=0\Longrightarrow{2\lambda\sin(x^2+y^2)=\display style\frac{4x-2y}{x-y}}$ (*)

Replacing with (*) in (1)

$4x+x \displaystyle\frac{4x-2y}{x-y}}=0\Rightarrow{4x(x-y)+x(4x-2y)=0}\Longrightarrow{x^2-4xy=0\longrightarrow{y=\displaystyle\frac{-x}{4}}}$

Replacing in (3)

$x^2+(\displaystyle\frac{-x}{4})^2=k\pi\Rightarrow{x=\pm{\sqrt[ ]{\displaystyle\frac{16}{17}k\pi}}}$

Well, I'm not sure if this is okey. If it is, how should I proceed?
Thanks for posting!

This problem is not really suited to the use of Lagrange multiplier techniques. The constraint $\cos (x^2+y^2)-1=0$ leads to a family of circles of radius $\sqrt{2k\pi}$ (don't forget the 2), for k = 0,1,2,...

On the first of those circles, when k=0 (actually a degenerate circle consisting of a single point at the origin), the function z takes its global minimum value 0. On the "k"-circle, z takes a local maximum value $4k\pi$ at the point $(\sqrt{2k\pi},0)$ , and a local minimum value $2k\pi$ at the point $(0,\sqrt{2k\pi})$ .

Not much of that information comes from the Lagrange method. In fact, at any point where the cosine function is equal to 1, the sine function vanishes. So equations (1) and (2) become x = y = 0, and the only extreme value that you get is the minimum at the origin.

If you want to use the Lagrange method, then I think you need to apply it separately to each constraint of the form $x^2+y^2 = 2k\pi$ .

**Ulysses** · November 5th 2010, 02:06 PM

I've just realized of it thinking geometrically. Thanks, this confirm my deduction. I've think of the elliptic paraboloid, and then I've realized that the contraint means that I'm looking for the extremes at the projection of this circles over the paraboloid, and then I realized that it happens when x=0 or y=0. And from there I get to the same conclusion than you. Any way, from lagrange it can be seen too from the first two equations. This holds, from (1) one can say that x=0 or $\lambda=0$
And from (2) y=0 or $\lambda=0$ , and from there I get this same conclusion, which I actually realized with a geometrical thinking.

And I forgot the 2 of course :P I'll correct it right now, didn't realized that it didn't holds for the odd values of k :O

Thanks!!

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