Having trouble figuring out lagrangian for this problem

[hellosoupy]

Hi!

This is a problem for which I have the answer in the back of the book, but I don't know how to get there.

Here is the problem:

A study conducted at a waste disposal site reveals soil contamination over a region that may be described roughly as the interior of the ellipse
[MATH]x^2/4 + y^2/9 = 1[/MATH]where x & y are in miles.

If the office at the site is at the point S(1,1), what is the radius of the smallest circle centered at S that contains the entire contaminated region?

Hint: The function
[MATH]f(x,y) = (x-1)^2 + (y-1)^2[/MATH]measures the square of the distance from S(1,1) to the point P(x,y). The required radius can be found by maximizing f(x,y) subject to a certain constraint.

My thought was to create the following Lagrangean:
[MATH]L=(x-1)^2 + (y-1)^2 + lambda*(36-9x^2-4y^2)[/MATH]
This would give
[MATH]Lx=2x-2-18x*lambda = 0[/MATH] and
[MATH]Ly=2y-2-8y*lambda = 0[/MATH]
This could be simplified to
[MATH]x(1-9*lambda) = 1[/MATH] and
[MATH]y(1-4*lambda) = 1[/MATH]
I tried to set them equal to each other and solve for y, but that gets me nowhere:
[MATH]y = x * (1-9*lambda) / (1-4*lambda)[/MATH]
Where have I gone astray?

Note: If it helps to work backwards, the answer is f(-0.49,-2.91) = 4.91 miles

Thanks in advance!

 

[JeffM]

Quick response.

Please check those answers. I may have made an arithmetic error, but I am not matching them.

Your third differential, which you are ignoring, does not contain lambda. Eliminate lambda from the first two differentials, and you will end up with two equations in x and y.

 

[hellosoupy]

Thanks, Jeff. Looking back, I think I was trying to just have a single x and a single y term, but that was a pretty dumb way to set up the equations.

I tried again but a) I think I made an algebra mistake (because I don't get 0 when I plug in the solution for x) and b) I don't know how to solve a fourth degree polynomial

These are practice problems before I take my quiz so I'm not married to finding a solution, but if you have any insights, please let me know!

 

[JeffM]

Longer response.

[MATH]L(x,\ y, \ \lambda) = (x - 1)^2 + (y - 1)^2 - \lambda (36 - 9x^2 - 4y^2).[/MATH]
We get three partial derivatives, NOT TWO. Pay attention to the whole system.

[MATH]\dfrac{\delta L}{\delta x} = 0 \implies 2x - 2 + 18x \lambda = 0 \implies 9x \lambda = 1 - x.[/MATH]
[MATH]\dfrac{\delta L}{\delta y} = 0 \implies 2y - 2 + 8y \lambda = 0 \implies y - 1 + 4y \lambda = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta \lambda} = 0 \implies 36 - 9x^2 - 4y^2 = 0 \implies x^2 = \dfrac{36 - 4y^2}{9}.[/MATH]
[MATH]y - 1 + 4y \lambda = 0 \implies 9xy - 9x + 4y (9x \lambda) = 0 \implies[/MATH]
[MATH]9xy - 9x + 4y(1 - x) = 0 \implies 5xy - 9x +4y = 0 \implies x(9 - 5y) = 4y.[/MATH]
[MATH]9 - 5y = 0 \implies y = \dfrac{9}{5} \implies x * 0 = \dfrac{36}{5} \implies 0 = \dfrac{36}{5}.[/MATH]
So 9 - 5y does not equal 0.

[MATH]x(9 - 5y) = 4y \implies x = \dfrac{4y}{9 - 5y} \implies x^2 = \dfrac{16y^2}{25y^2 - 90y + 81}. [/MATH]
[MATH]\therefore \dfrac{16y^2}{25y^2 - 90y + 81} = \dfrac{36 - 4y^2}{9}.[/MATH]
Now that leaves a quartic to be solved. The book’s solution does not fit exactly. So probably I made an algebraic or arithmetic error along the way. But the calculus part of the problem is straight forward.

 

[JeffM]

When I put that mess into an equation solver, I got y is approximately -2.907.

So it does agree with the book. The book's approximation threw me off.

(There is another solution for y, but it is positive and the correct value for x and y must be in the third quadrant.)

 

[hellosoupy]

I think I might be losing my mind, but how did you get from

y−1+4yλ=0
to
9xy−9x+4y(9xλ)=0

?

Also, my textbook has us adding the lambda term. Thoughts?

 

[JeffM]

[MATH]y - 1 + 4y \lambda = 0 \implies 9x(y - 1 + 4y \lambda) = 9x * 0 \implies 9xy - 9x + 9x(4y \lambda) = 0 \implies 9xy + 4y(9x \lambda) = 9x.[/MATH]
Mechanically, you are just solving a system of simultaneous equations. You were taught long ago that there are multiple methods to solve such systems when the equations are linear. The same methods work if the equations are non-linear although with non-linear equations you may get multiple solutions.

 

[hellosoupy]

Thank you for working through that! Strange that when I went through your steps with my textbook's way of doing it (adding the lambda term) I got to the same final equation!