Find the points closest to the origin

[NewtonianAlch]

Homework Statement

Find the points $x^2 + xy + y^2 = 2$ that are closest to the origin.

Homework Equations

Distance Formula

The Attempt at a Solution

I have to first solve this without using Lagrange Multipliers.

This is essentially an ellipse. So I first completed the square:

$3/4\,{x}^{2}+ \left( y+1/2\,x \right) ^{2}=2$

I was thinking first I should separate $x$ and $y$, and use that in the distance formula, but I can't seem to isolate $y$, is my approach wrong?

 

[Mark44]

Homework Statement

Find the points $x^2 + xy + y^2$ that are closest to the origin.

Shouldn't this be an equation?

Homework Equations

Distance Formula

The Attempt at a Solution

I have to first solve this without using Lagrange Multipliers.

This is essentially an ellipse. So I first completed the square:

$3/4\,{x}^{2}+ \left( y+1/2\,x \right) ^{2}=2$

Where did the 2 come from? Are you starting with $x^2 + xy + y^2 = 2$?

I was thinking first I should separate $x$ and $y$, and use that in the distance formula, but I can't seem to isolate $y$, is my approach wrong?

 

[NewtonianAlch]

Shouldn't this be an equation?
Where did the 2 come from? Are you starting with $x^2 + xy + y^2 = 2$?

Yes, sorry. That was silly of me, I edited the original post to correct that error.

 

[Mark44]

The equation y² + xy + x² - 2 = 0 is quadratic in y, so you can use the Quadratic Formula to solve for y. (You'll get two values with the ±, meaning there are two functions of x.)

Your problem is to find the minimum value of D(x) = √(x² + y²). Here you will need to substitute for y from the previous work, making D really a function of x alone.

Alternatively, you could find the minimum value of D²(x) = x² + y², again making the substitutions for y. Since there are two functions, you'll need to do the work for both of them. I don't know if there is any symmetry you can exploit to save work. The graph is an ellipse, but one that has been rotated.

 

[NewtonianAlch]

Thanks for that, I will look into it.

 

[Ray Vickson]

Homework Statement

Find the points $x^2 + xy + y^2 = 2$ that are closest to the origin.

Homework Equations

Distance Formula

The Attempt at a Solution

I have to first solve this without using Lagrange Multipliers.

This is essentially an ellipse. So I first completed the square:

$3/4\,{x}^{2}+ \left( y+1/2\,x \right) ^{2}=2$

I was thinking first I should separate $x$ and $y$, and use that in the distance formula, but I can't seem to isolate $y$, is my approach wrong?

As an alternative method, you could solve the problem as a constrained optimization, using a Lagrange multiplier approach. You want to minimize f(x,y) = x^2 + y^2 (the square of the distance) subject to g(x,y) = 0, where g(x,y) = x^2 + y^2 + x*y-2.

RGV

 

[Mark44]

As an alternative method, you could solve the problem as a constrained optimization, using a Lagrange multiplier approach.

In post #1 it says

I have to first solve this without using Lagrange Multipliers.

You want to minimize f(x,y) = x^2 + y^2 (the square of the distance) subject to g(x,y) = 0, where g(x,y) = x^2 + y^2 + x*y-2.

 

[D H]

The equation y² + xy + x² - 2 = 0 is quadratic in y, so you can use the Quadratic Formula to solve for y. (You'll get two values with the ±, meaning there are two functions of x.)

Another option is to use polar coordinates. The equation of the ellipse becomes $r^2(1+\sin\theta\cos\theta) = 2$, where $r$ is the distance from the origin.

And that's a big enough hint at this point.

 

[Ray Vickson]

In post #1 it says

I missed that remark about no Lagrange multipliers, although I cannot see why anyone would say "you cannot use such-and-such a technique". What's the point of that? However, the OP must live with it.

RGV

 

[D H]

It's probably because the next part of the problem (not shown here) was to solve the same problem, but this time using Lagrange multipliers.