Giải thích các bước giải:

$\begin{aligned}
& \text{Gọi } \alpha \text{ là góc tạo bởi chiếc thang và mặt đất } \left(0 < \alpha < \dfrac{\pi}{2}\right) \\
& \text{Chiều dài chiếc thang là } L(\alpha) \\
& L(\alpha) = \dfrac{12}{\cos\alpha} + \dfrac{10}{\sin\alpha} \\

&\text{Ta có:}\\
& L'(\alpha) = \dfrac{12\sin\alpha}{\cos^2\alpha} - \dfrac{10\cos\alpha}{\sin^2\alpha} \\


&\text{Để chiều dài ngắn nhất:}\\

& L'(\alpha) = 0 \\
& \dfrac{12\sin\alpha}{\cos^2\alpha} = \dfrac{10\cos\alpha}{\sin^2\alpha} \\
& 12\sin^3\alpha = 10\cos^3\alpha \\
& \tan^3\alpha = \dfrac{10}{12} = \dfrac{5}{6} \\
& \tan\alpha = \sqrt[3]{\dfrac{5}{6}} \\
& \cos^2\alpha = \dfrac{1}{1+\tan^2\alpha} = \dfrac{1}{1+\left(\dfrac{5}{6}\right)^{\dfrac{2}{3}}} = \dfrac{6^{\dfrac{2}{3}}}{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}} \\
& \sin^2\alpha = \dfrac{\tan^2\alpha}{1+\tan^2\alpha} = \dfrac{\left(\dfrac{5}{6}\right)^{\dfrac{2}{3}}}{1+\left(\dfrac{5}{6}\right)^{\dfrac{2}{3}}} = \dfrac{5^{\dfrac{2}{3}}}{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}} \\

&\text{Khi đó:}\\
& L = \dfrac{12}{\sqrt{\dfrac{6^{\dfrac{2}{3}}}{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}}}} + \dfrac{10}{\sqrt{\dfrac{5^{\dfrac{2}{3}}}{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}}}} \\
& L = 12 \dfrac{\sqrt{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}}}{6^{\dfrac{1}{3}}} + 10 \dfrac{\sqrt{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}}}{5^{\dfrac{1}{3}}} \\
& L = \sqrt{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}} \left( \dfrac{12}{6^{\dfrac{1}{3}}} + \dfrac{10}{5^{\dfrac{1}{3}}} \right) \\
& L = \sqrt{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}} \left( 2 \cdot 6^{\dfrac{2}{3}} + 2 \cdot 5^{\dfrac{2}{3}} \right) \\
& L = 2 \left( 6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}} \right) \sqrt{6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}}} \\
& L = 2 \left( 6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}} \right)^{\dfrac{3}{2}} \\
& L = \left( 2^{\dfrac{2}{3}} \left( 6^{\dfrac{2}{3}}+5^{\dfrac{2}{3}} \right) \right)^{\dfrac{3}{2}} \\
& L = \left( 12^{\dfrac{2}{3}} + 10^{\dfrac{2}{3}} \right)^{\dfrac{3}{2}} \\
& L \approx 31,1 \\
& \text{Kết quả: Độ dài ngắn nhất của chiếc thang là } 31,1 \text{ m}
\end{aligned}$