Maximinzing and minimizing a circle in an ellipse using lagrange multipliers Consider the following question:

The equation $73x^{2}$ + $72xy$ + $52y^{2} = 100$ defines an ellipse which is
  centered at the origin, but has been rotated about it. Find the
  semiaxes of this ellipse by maximizing and minimizing $f(x, y) = x^{2} +
y^{2}$ on it.

This problem seems very complicated to me. I'm self-studying multivariable calculus and I cannot figure out this problem. Let $g(x, y, z) = 73x^2 + 72xy + 52y^2 - 100$ and $f(x, y, z) = x^2 + y^2$
First, I computed $$g_{x} = 146x + 72y$$
$$g_{y} = 104y + 72x$$
$$g_{z} = 0 $$
$$f_{x} = 2x$$
$$f_{y} = 2y.$$
$$f_{z} = 0.$$
Then, I have
$$2x = \lambda (146x + 72y)$$
$$2y = \lambda(104y + 72x)$$
$$73x^{2} + 72xy + 52y^{2} = 100$$
I don't know how to proceed since this is a system of three variables. Also, this is my first lagrange multiplier attempt so I'm not confident about what to do next either. I've been using Paul's Online Notes to try and understand. My ultimate goal is to be able to apply this on inequalities because I've heard lagrange multipliers are helpful in math olympiad. This seemed like a good example. 
 A: This is a homogeneous problem and can be handled easily. Making $y = \mu x$ we have.
$$
\begin{cases}
x^2+y^2 = x^2(1+\mu^2)\\
73x^2+72 x y + 52 y^2=x^2(73+72\mu+52\mu^2) = 100
\end{cases}
$$
so the problem is equivalent to 
$$
\min_{\mu}f(\mu) = \frac{100(1+\mu^2)}{73+72\mu+52\mu^2}
$$
now 
$$
f'(\mu)=0\equiv (1+\mu)(16\mu-37) = 0
$$
and thus we have $\mu = -1$ and $\mu = \frac{16}{37}$ as stationary points etc.
A: Lagrange multiplier, in essence, says the gradient of a function $f(x_1,x_2..x_n)$ being maximized or minimized subjected to a constraint $g(y_1, y_2...y_m)$ will be such that, at the critical points $\triangledown f(x_1,x_2..x_n)=\lambda \triangledown g(y_1, y_2...y_n)$. 
You have the equation set up wrong, In your question the constraint is $73x^2+72xy+52y^2=100=g(x,y)$ and the function you are maximizing is $x^2+y^2=f(x,y)$.
So your equation then becomes $\triangledown(x^2+y^2)=\lambda\triangledown(73x^2+72xy+52y^2=100)$. after solving it you will have two equations and 3 unknowns; however, your third equation will be $73x^2+72xy+52y^2=100$.
$$f_x=\lambda g_x$$
$$f_y=\lambda g_y$$
$$73x^2+72xy+52y^2=100$$
Solve the equations for $x,y$, and $z$. That is the point where your function will reach its maximum and minimum. In this particular case, though, I believe you will get more than one max and min point. I am not sure, you will have to solve it.
A: The problem we have is:
$$ \underset{x,y}{\text{minimize}}~ x^2+y^2 \\ s.t.~73x^2+72xy+52y^2=100$$
We first find the gradients for $f$ and $g$: 
$$\nabla f = (2x,~2y)\\ \nabla g= (146x+72y,~72x+104y)$$. 
The Lagrange multipliers method tells us that $\nabla f= \lambda \nabla g$. This gives us the following two equations:
$$2x =\lambda(146x+72y) ~(1) \\ 2y=\lambda(72x+104y)~(2)$$
This is a system of 2 equations with 3 variables. In order to solve it, we need a third equation. We know that the solution must lie on the ellipse, this gives us the last equation:
$$73x^2+72xy+52y^2=100 ~(3)$$
The problem is now to solve this system of equations.
Finding $\lambda$ in $(1)$ and replacing in $(2)$, and simplifying we get the following:
$$12x^2-7xy-12y^2=0$$
This equation has two solutions, which are straight lines:
$$x=\dfrac{7y \pm 25y}{24}$$ and those are the semiaxes of the ellipse. 
The points where these lines intercept the ellipse are the solutions to the optimisation problem. The points are $(0.8,0.6)$ and $(-0.8,-0.6)$ for the minimization problem and $(-1.2,1.6),~(1.2,-1.6)$ for the maximization one.  
