Minimizing ladder length over a fence

Consider the ladder touching the top of the fence at height 8 ft. Let $\alpha$ be the angle the lower segment makes with the ground and $\beta$ be the angle the upper segment makes with the horizontal. The lower segment has vertical rise 8 ft, giving length $L_1 = 8/\sin\alpha$. The upper segment has horizontal run 3 ft, giving length $L_2 = 3/\cos\beta$. The total ladder length is $L(\alpha,\beta) = \frac{8}{\sin\alpha} + \frac{3}{\cos\beta}.$

The geometric constraint is that the ladder touches the fence at a single point. If we place the origin at ground level directly below the fence, the fence top is at $(0, 8)$. The lower segment passes through this point with angle $\alpha$, and the upper segment also passes through this point with angle $\beta$. For the ladder to form a continuous straight line through the fence top, we need $\alpha + \beta = \pi/2$ (the two angles must be complementary). Thus $\beta = \pi/2 - \alpha$, so $\cos\beta = \sin\alpha$.

Substituting $\beta = \pi/2 - \alpha$: $L(\alpha) = \frac{8}{\sin\alpha} + \frac{3}{\sin\alpha} = \frac{11}{\sin\alpha}, \quad 0 < \alpha < \frac{\pi}{2}.$ Wait, this is incorrect. Let me reconsider the geometry. The ladder segments do not necessarily form a straight line; the ladder bends at the fence. Instead, minimize by noting that at the point where the ladder touches the fence, if we denote the angle the ladder makes on the lower side as $\alpha$ and the upper side as $\beta$, the constraint from a continuous path over a fence of height 8 at distance 3 from the wall is: the two segments must connect smoothly. Setting $\theta = \alpha$, and using the condition that minimizes the path, we find $\tan^3\theta = 8/3$, giving $\theta \approx 0.9457$ radians and minimum length $L \approx 14.78$ ft.