<s>[INST] Determining the minimal length of the hypotenuse How can I solve the following problem?



The hint I received is that the minimal length is not given by $L=15$. I feel like I need more information to solve this problem because I can't get any further. Any hints, tips, solutions are appreciated!
Thank you.
 [/INST] Alternatively, denote the hypotenuse of small right-angle triangle by $z$. Then:
$$\frac{L}{z}=\frac{8+x}{x} \ \ \text{and} \ \ z^2=x^2+3^2 \Rightarrow\\
L=\left(\frac{8}{x}+1\right)z \to min, \ \ s.t. \ \ z=\sqrt{x^2+9}.$$
Subsituting:
$$L(x)=\left(\frac{8}{x}+1\right)\sqrt{x^2+9}, \\
L'(x)=\left(-\frac{8}{x^2}\right)\sqrt{x^2+9}+\left(\frac 8x+1\right)\frac{x}{\sqrt{x^2+9}}=0 \Rightarrow \\
\frac{-8(x^2+9)+(8+x)x^2}{x^2\sqrt{x^2+9}}=0 \Rightarrow x^3=72 \Rightarrow x=2\sqrt[3]{9}.$$
Hence:
$$\tan \alpha = \frac 3x \Rightarrow \alpha = \arctan \frac{3}{2\sqrt[3]{9}}=0.624763... \ \text{rad}=35.8^\circ.\\
L(2\sqrt[3]{9})=14.99216...$$
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