 Solution

Using $J=1$ and the preceding first Piola-Kirchhoff stress tensor,Its eigenvalues are $\mu (1+\sqrt{5})/2$, $\mu (1-\sqrt{5})/2$, and zero. Because it is symmetric, its largest traction magnitude on a unit spatial normal is the largest absolute eigenvalue. Put $\phi =(1+\sqrt{5})/2$. The positive eigenvalue has the strictly largest modulus, with normalThe question asks for the plane in the reference configuration, so this spatial normal must be pulled back. Nanson's formula gives $N$ parallel to $F^{T}n$, andTherefore the required plane and maximum areEither sign of its unit normal describes the same plane. Maximizing over reference normals is equivalent to maximizing over spatial normals because the map $N\mapsto F^{-T}N/|F^{-T}N|$ is a bijection of unit normals.