I want to minimize $f(\mathbf{x})$ where $\mathbf{x} \in \mathbb{R}^n $ and
$$ f(\mathbf{x}) = \frac{1}{2} \mathbf{x}^T \mathbf{A x} + \mathbf{b}^T \mathbf{x} $$
on the hyperplane given by $\mathbf{n}^T (\mathbf{x} - \mathbf{p}) = 0 $
It is given that $\mathbf{A}$ is positive definite. I'd like to find the minimizing vector $\mathbf{x}$ and the minimum value of $f$.
Using Lagrange multiplier method, function $g(\mathbf{x}, \lambda) $ is defined as follows
$ g = \frac{1}{2} \mathbf{x}^T \mathbf{A x} + \mathbf{b}^T \mathbf{x} + \lambda (\mathbf{n}^T (\mathbf{x} - \mathbf{p}) ) $
Then at the minimum, the following holds:
$ \mathbf{\nabla_x} g = \mathbf{0} \hspace{36pt} (1)$
$ \mathbf{n}^T (\mathbf{x} - \mathbf{p}) = 0 \hspace{36pt} (2)$
Now,
$ \mathbf{\nabla_x} g = \mathbf{Ax} + \mathbf{b} + \lambda \mathbf{n} \hspace{36pt} (3) $
It follows that at the minimum,
$ \mathbf{x} = - \mathbf{A}^{-1} ( \lambda \mathbf{n} + \mathbf{b} ) \hspace{36pt} (4) $
Substituting $(4)$ into $(2)$:
$ \lambda = -\dfrac{ \mathbf{n}^T (\mathbf{A}^{-1} \mathbf{b} + \mathbf{p} ) }{\mathbf{n}^T \mathbf{A}^{-1} \mathbf{n} } \hspace{36pt} (5)$
With $(5)$ and $(4)$ the minimizing $\mathbf{x}$ is now known. Substituting it into the expression for $f(\mathbf{x})$ yields the minimum value of $f$.