I am consider the following problem:
$$ \max_{x\in\mathbb{R}^n} V(x) = x^\intercal P x, \qquad \text{s.t.} \quad |b^\intercal x| \leq r, $$
where $P > 0$ is positive definite. First, let's consider the scalar case:
$$ \max_{x\in\mathbb{R}} V(x) = px^2, \qquad \text{s.t} \quad |bx| \leq r $$
If you draw this out, it is clear that the maximum occurs when equality is achieved, i.e. $x^* = \pm r/b$, which gives $V^* = V(x^*) = pr^2/b^2$. However, now I want to generalize this to the case where $x\in\mathbb{R}^n$. By analogy, the maximum will occur when $b^\intercal x^* = r$, however I am not sure how to plug this into $V(x)$ to get $V^*$.
you may want to consider the projection matrix $S := I - \frac{1}{\mathbf b^T \mathbf b}\mathbf b\mathbf b^T$
$S P S = \big(P^\frac{1}{2}S\big)^T\big( P^\frac{1}{2}S\big)\succeq 0$
and in fact has rank $n-1$
Then subject to $\big \vert\mathbf b^T \mathbf x\big \vert\leq r$
$\max_{x\in\mathbb{R}^n} V(x) = \max_{x\in\mathbb{R}^n} \mathbf x^T P \mathbf x \geq \max_{(x \perp b) \in\mathbb{R}^n} V(x) =\max_{x\in\mathbb{R}^n} \mathbf x^T S P S \mathbf x = \infty$
Put differently this maximizaiton problem is unbounded and the max doesn't exist.