For the box on the verge of moving, the net force is zero. The static friction force is at its maximum, $f_s = \mu_s N$.

Sum of vertical forces:

$$ \sum F_y = N + T_{max} \sin\theta - W = 0 \implies N = W - T_{max} \sin\theta $$

Sum of horizontal forces:

$$ \sum F_x = T_{max} \cos\theta - f_s = 0 \implies T_{max} \cos\theta = \mu_s N $$

Substitute the expression for the normal force $N$ into the horizontal force equation:

$$ T_{max} \cos\theta = \mu_s (W - T_{max} \sin\theta) $$

Solve for the weight $W$:

$$ \mu_s W = T_{max} \cos\theta + \mu_s T_{max} \sin\theta $$ $$ W(\theta) = \frac{T_{max}}{\mu_s} (\cos\theta + \mu_s \sin\theta) $$

To maximize $W$, we must maximize the function $f(\theta) = \cos\theta + \mu_s \sin\theta$. This expression is of the form $A\cos\theta + B\sin\theta = R\cos(\theta - \alpha)$, where $R = \sqrt{A^2+B^2}$ and $\tan\alpha = B/A$. Here, $A=1$ and $B=\mu_s$. Thus $R = \sqrt{1+\mu_s^2}$ and $\tan\alpha = \mu_s$.

The function $f(\theta)$ has a maximum value of $R = \sqrt{1+\mu_s^2}$, which occurs when $\theta = \alpha$. Therefore, the optimal angle is given by $\tan\theta = \mu_s$.

The maximum weight $W_{max}$ is found by substituting the maximum value of $f(\theta)$ into the equation for $W$:

$$ W_{max} = \frac{T_{max}}{\mu_s} ( \sqrt{1+\mu_s^2} ) $$