Constrained Extremes of z=2x^2+y^2 over Cosine Circles

[Telemachus]

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): $$4x-2y+(2\lambda \sin(x^2+y^2))(x-y)=0\Longrightarrow{2\lambda\sin(x^2+y^2)=\displaystyle\frac{4x-2y}{x-y}}$$ (*)

Replacing 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?

 

[micromass]

On the first glance, it seems alright.

So you have found the candidate extremes $$\pm{\sqrt[ ]{\displaystyle\frac{16}{17}k\pi}}}$$. Now you will all have to fill them in in the function 2x²+y² to see which ones are the real extrema.

 

[Telemachus]

But it will depends on "k". How do I handle it?

Thx for posting micromass ;)

 

[micromass]

Well, k is just a parameter. For every k we have a value. So letting k:1,2,3,... we have

$$\pm{\sqrt[ ]{\displaystyle\frac{16}{17}\pi}}},\pm{\sqrt[ ]{\displaystyle\frac{16}{17}2\pi}}},\pm{\sqrt[ ]{\displaystyle\frac{16}{17}3\pi}}},...$$

So you will have an infinity of values you'll have to plug in the function.

Just try to figure out for which k 2x²+y² becomes maximal...

Hope that helped...

 

[Telemachus]

You mean I should consider when k is pair and when its odd?

 

[micromass]

Maybe it's simpler.

You wrote:
Replacing in (1): then y=-x/4

But what happens if we also replace in (2)? Then

$$2y+y\frac{4x-2y}{x-y}=0~\Rightarrow~ 2y(x-y)+y(4x-2y)=0~\Rightarrow~6xy-4y^2~\Rightarrow~3x=4y$$

This is another constraint on x and y. Together with y=-x/4, this should limit the cases...

 

[Telemachus]

mm I think what I did is wrong. I've been thinking on how it looks, and the graph of z is an elliptic paraboloid, and the restrictions are circles. Then I've realized that the minimum would be when x=0, and the maximum when y=0, because what we are looking are the maximums on the projections of this circles on the paraboloid, its obvious as the function is "larger" on the x-axis that there would be a minimum there, and a maximum on the y-axis for every circle, and the absolute minimum when k=0=x=y. I think this is the right answer, and from the equations above one can see that 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.