Rayleigh quotient and eigenvalues

Lagrange. Maximize $f(x) = x^T A x$ under $g(x) = x^T x = 1$. The gradients are
$$\nabla f = 2Ax, \qquad \nabla g = 2x.$$
The condition $\nabla f = \lambda \nabla g$ gives
$$Ax = \lambda x.$$
So $x$ is an eigenvector of $A$ associated with the eigenvalue $\lambda$.

Value at a critical point. If $Ax = \lambda x$, then
$$f(x) = x^T A x = x^T (\lambda x) = \lambda x^T x = \lambda.$$
So the maximum of $f$ on the sphere is the largest eigenvalue $\lambda_{\max}$.

Rayleigh quotient. For $x \neq 0$, $R(x) = f(x/\|x\|)$, so the maximum of $R$ over $\mathbb{R}^n \setminus \{0\}$ is also $\lambda_{\max}$.

Conclusion. The maximum is attained at a unit eigenvector associated with $\lambda_{\max}$.