= Solution
The insulated boundary imposes the <Neumann boundary condition> $T_{x_1}(0,x_2,t)=0$. Use the <method of images> by reflecting the initial data evenly across $x_1=0$. At the origin, the original sector and its reflection contribute equally. Using the given two-dimensional <heat kernel>,
$$
T(0,0,t)=\frac{2T_0}{4\pi Kt}\int_{-\pi/4}^{\pi/4}\int_1^2e^{-r^2/(4Kt)}r\,dr\,d\theta
=\frac{T_0}{2}\left(e^{-1/(4Kt)}-e^{-1/(Kt)}\right).
$$
Differentiation gives
$$
\frac{dT}{dt}=\frac{T_0}{8Kt^2}\left(e^{-1/(4Kt)}-4e^{-1/(Kt)}\right).
$$
This vanishes exactly when $e^{3/(4Kt)}=4$. Its sign is positive before that time and negative afterwards. Moreover $T\to0$ as $t\downarrow0$ and as $t\to\infty$. Therefore this is the unique global maximum:
$$
\boxed{t_{\max}=\frac3{8K\log2},\qquad T_{\max}=\frac{3T_0}{8\,4^{1/3}}}.
$$
The factor two from the insulated-boundary reflection is essential; applying the full-plane <heat kernel> only to the unreflected sector would give half the temperature.
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