circle inside a function

[Mimmo]

For what values \(\displaystyle f(x,y) = x^2 + y^2\) do we have \(\displaystyle x^2+y^2-xy \leq 1\)?

If we set x = 0 we get the circle with the radius 1. Is this the answer?

\(\displaystyle x^2+y^2-xy \leq 1\)
\(\displaystyle x^2+y^2-1 \leq xy\)
\(\displaystyle 1-1 \leq xy\)
\(\displaystyle 0 \leq xy\)

This isn't true for the part of the circle in the second and fourth quadrant! What's wrong? Please help.

 

[arthur ohlsten]

I believe its a circle of radius 2
Are you studying rotation of axis to get rid of xy terms?
make the substitution

x' = x cos @ -y sin @
y' = x sin @+y cos @
let @ = pi/4

Arthur

 

[Gene]

Well, I have to disagree with Art.
If you solve for y in x²+y²+2xy=1 (by completing the square) you get the unfriendly
y=x/2+sqrt(1-3x²/4)
Then (worse and worse)
r²=x²+y² gives the radius of a circle thru f(x,y), which you have to minimize. "All" you have to do is find the derivitive of
sqrt(x/2+sqrt(1-3x²/4))²+x²)

It can be simplified by finding the derivitive of r² which lets you drop the sqrt.

Perhaps there is a trig substitution that would make it "friendly" but I don't see it.

If you are willing to recognize the equation as an elipse rotated 45° you can solve for y=-x as the minimum, but that doesn't seem fair.

Any way you do it, you should end up with
r=.8045302465

 

[pka]

This is what I found.
In the first and third quadrants it is \(\displaystyle \L
\{ (x,y):x^2 + y^2 \le 1\}\).

In quadrant II it is \(\displaystyle \L
\left\{ {(x,y):y \le \frac{{x + \sqrt {4 - 3x^2 } }}{2}} \right\}\).

In quadrant IV it is \(\displaystyle \L
\left\{ {(x,y):y \ge \frac{{x - \sqrt {4 - 3x^2 } }}{2}} \right\}\)

 

[Mimmo]

Yes, it should be a type of elipse.

I've also tried the lagrange (not exactly) method.

\(\displaystyle F(x,y) = x^2+y^2\)
\(\displaystyle G(x,y)=x^2+y^2-xy-1 = 0\)

\(\displaystyle \nabla F = (2x,2y)\)
\(\displaystyle \nabla G = (2x-y,2y-x)\)

\(\displaystyle 0 = \det (\nabla F, \nabla G) = -2x^2 + 2y^2 \Rightarrow x = +- y\)

By plugging this in G I get \(\displaystyle x = y \Rightarrow x = 1, x = -y \Rightarrow x = 1/\sqrt3\)

Neather one of them gives me .8045302465 :\

 

[Gene]

My finger must have slipped. Now I get .8164966536 which I wouldn't have recognized as sqrt(2/3) if I had gotten it right.
------------------
Gene

 

[Mimmo]

Then my solution was correct (x = 1/sqrt3)

But, I also got x = 1 as an alternatively solution. How do I show that this solution, algebraically, does NOT work and that x = 1/sqrt3 is the solution?

 

[Gene]

x=1 is the solution for the longer axis where dr/dx = zero because it is at the maximum radius of a doubly tangent circle. It gives the circle outside the elipse, not inside. x=1/sqrt(3) gives the smallest doubly tangent circle, inside the elipse.
---------------
Gene
PS I figured out where I went wrong with .804... My TI-83 changes the x value (a little) when you move vertically between two curves. Don't know why.