Minimizing ladder length over a fence

Rewrite $L(\theta) = 4\sin^{-1}\theta + 5\cos^{-1}\theta$ (using shorthand for $(\sin\theta)^{-1}$ etc.) and differentiate: $L'(\theta) = -\frac{4\cos\theta}{\sin^2\theta} + \frac{5\sin\theta}{\cos^2\theta}.$ Setting $L'(\theta)=0$ gives $\frac{5\sin\theta}{\cos^2\theta} = \frac{4\cos\theta}{\sin^2\theta}$, which simplifies to $5\sin^3\theta = 4\cos^3\theta$. Dividing both sides by $\cos^3\theta$ yields $\tan^3\theta = \frac{4}{5}$.

Solving gives $\theta = \arctan\left(\left(\frac{4}{5}\right)^{1/3}\right)$. Since $L(\theta) \to \infty$ as $\theta \to 0^+$ or $\theta \to \frac{\pi}{2}^-$, this critical point must be the minimum. At the critical point where $\tan^3\theta = \frac{4}{5}$, we have $\sin\theta = \frac{(4/5)^{1/3}}{\sqrt{1 + (4/5)^{2/3}}}$ and $\cos\theta = \frac{1}{\sqrt{1 + (4/5)^{2/3}}}$. The minimum ladder length is $L_{\min} = \frac{4}{\sin\theta} + \frac{5}{\cos\theta} = 4\sqrt{1 + (4/5)^{2/3}} + 5\sqrt{1 + (4/5)^{2/3}} = 9\sqrt{1 + (4/5)^{2/3}} \approx 12.7 \text{ ft}.$ Numerically, $\theta \approx 0.727$ radians.

The ladder length increases without bound near both endpoints. As $\theta$ approaches zero, the term $\frac{4}{\sin\theta}$ blows up, and as $\theta$ approaches $\frac{\pi}{2}$, the term $\frac{5}{\cos\theta}$ blows up. This is exactly why the critical point found by setting the derivative to zero must give a minimum rather than a maximum. The function's behavior at the boundaries confirms the interior critical point is the global minimum on the open interval.