= Solution
{c}
Using $J=1$ and the preceding <first Piola-Kirchhoff stress tensor>,
$$
T=T_RF^T=\mu\begin{pmatrix}1&1&0\\1&0&0\\0&0&0\end{pmatrix}.
$$
Its <eigenvalues> are $\mu(1+\sqrt5)/2$, $\mu(1-\sqrt5)/2$, and zero. Because it is <symmetric>, its largest <traction> magnitude on a unit spatial <normal> is the largest absolute <eigenvalue>. Put $\varphi=(1+\sqrt5)/2$. The positive <eigenvalue> has the strictly largest modulus, with <normal>
$$
n=\pm\frac{(\varphi,1,0)}{\sqrt{\varphi^2+1}}.
$$
The question asks for the <plane> in the <reference configuration>, so this spatial <normal> must be pulled back. <Nanson's formula> gives $N$ parallel to $F^Tn$, and
$$
F^T(\varphi,1,0)^T=(\varphi,\varphi+1,0)^T
=\varphi(1,\varphi,0)^T.
$$
Therefore the required <plane> and maximum are
$$
\boxed{x_1+\varphi x_2=0,\qquad
\max|s|=\mu\varphi=\frac{\mu(1+\sqrt5)}2.}
$$
Either 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.
\Image[/past-exam-of-the-solid-mechanics-course-of-the-university-of-oxford/2007-shear-tractions.png]
{title=Reference and spatial normals differ when maximizing the Cauchy <traction>}
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