- #1
Telemachus
- 835
- 30
I have some trouble with this exercise. It says as follows: Find the constrained extremes for [tex]z=2x^2+y^2[/tex] over [tex]\cos (x^2+y^2)-1=0[/tex]
I think that difficult of this problem is that I have a family of circles that satisfies [tex]\cos (x^2+y^2)-1=0[/tex], the family of circles with radius [tex]r=\sqrt[ ]{k \pi}[/tex].
This is what I did.
[tex]\cos (x^2+y^2)-1=0\Rightarrow{x^2+y^2=k \pi},k\in{Z}[/tex]
[tex]F(x,y,\lambda)=2x^2+y^2-\lambda(cos(x^2+y^2)-1)=0[/tex]
[tex]\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}[/tex]
From (1) and (2): [tex]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}}[/tex] (*)
Replacing in (1)
[tex]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}}}[/tex]
Replacing in (3)
[tex]x^2+(\displaystyle\frac{-x}{4})^2=k\pi\Rightarrow{x=\pm{\sqrt[ ]{\displaystyle\frac{16}{17}k\pi}}}[/tex]
Well, I'm not sure if this is okey. If it is, how should I proceed?
I think that difficult of this problem is that I have a family of circles that satisfies [tex]\cos (x^2+y^2)-1=0[/tex], the family of circles with radius [tex]r=\sqrt[ ]{k \pi}[/tex].
This is what I did.
[tex]\cos (x^2+y^2)-1=0\Rightarrow{x^2+y^2=k \pi},k\in{Z}[/tex]
[tex]F(x,y,\lambda)=2x^2+y^2-\lambda(cos(x^2+y^2)-1)=0[/tex]
[tex]\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}[/tex]
From (1) and (2): [tex]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}}[/tex] (*)
Replacing in (1)
[tex]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}}}[/tex]
Replacing in (3)
[tex]x^2+(\displaystyle\frac{-x}{4})^2=k\pi\Rightarrow{x=\pm{\sqrt[ ]{\displaystyle\frac{16}{17}k\pi}}}[/tex]
Well, I'm not sure if this is okey. If it is, how should I proceed?