Relative eigenvalues and the Rayleigh quotient in tensor notation I'm working through Pavel Grinfeld's Introduction to Tensor Analysis and the Calculus of Moving Surfaces and I'm very stuck on exercise 118, which reads:

Show that the eigenvalues of the generalized equation (7.71) are given by the Rayleigh quotient
  $$\lambda = A_{ij} x^i x^j.$$

This references an equation from a preceding section of text:

In this section, we show that, much like $A x = b$, the eigenvalue problem
  $$A x = \lambda M x\tag{7.66}$$
  can be formulated as a variational problem. The matrix $M$ is assumed to be symmetric and positive define [sic]. The variational formulation is to find the extrema of
  $$f(x) = A_{ij} x^i x^j\tag{7.67}$$
  subject to the constraint that
  $$M_{ij} x^i x^j = 1.\tag{7.68}$$
  Geometrically, equation (7.68) states that the vector $x^i$ unit length [sic]. Use a Lagrange multiplier $\lambda$ to incorporate the constraint on the augmented function $E(x, \lambda)$
  $$E(x, \lambda) = A_{ij} x^i x^j - \lambda (M_{ij} x^i x^j - 1).\tag{7.69}$$
  Following earlier analysis,
  $$\frac{1} {2} \frac{\partial E(x)} {\partial x^i} = A_{ij} x^j - \lambda M_{ij} x^j.\tag{7.70}$$
  Equating the partial derivatives to zero yields
  $$A_{ij} x^j = \lambda M_{ij} x^j\tag{7.71}$$
  which is equivalent to the eigenvalue problem (7.66).

First I tried plugging the expression for $\lambda$ (with renamed indices) into the RHS of equation (7.71) to try and obtain the LHS, but I didn't really get any further than plugging it in. Then I considered solving the variational problem for $\lambda$ supposing we'd found an $x$ that works, but had no idea where to begin.

You can just multiply both sides of $(7.71)$ by $x^i$ and use the constraint:
$$ A_{ij} x^j x^i = \lambda M_{ij} x^i x^j = \lambda  \,.$$
