Let $A=\lim_{r \rightarrow +\infty} \frac{Vol(B(o,r))}{\omega_{n} r^{n}}$ for any Riemannian manifold $(\mathbb{M}^{n},g)$ with nonnegative Ricci curvature. Here $\omega_{n}$ is the volume of unit ball in $\mathbb{R}^n$. We call $A$ the cone angle at infinity or asymptotic volume ratio, and the manifold is cone-like if $A>0$. I got a direct question with this definition. Let $(\mathbb{M}^{n},g)$ be a metric product of two manifolds $(\mathbb{M}_1^{k},g_1,A_1)$ and $(\mathbb{M}_2^{n-k},g,A_2)$ both with nonnegative curvature and cone-like. Does the cone angle of the product manifold only depends on $k$, $n$, $A_1$ and $A_2$. It sounds like the new cone angle will be some average of $A_1$ and $A_2$, but I failed to get a clean formula even assuming product of two rotationally symmetric cones in $R^3$ (like surface of evolution). Is there something simple I miss? Thanks!