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.
 A: Hint: use that $$\tan(\alpha)=\frac{3}{x}$$ and $$\cos(\alpha)=\frac{8+x}{L}$$
$$L'(\alpha)=\frac{8(1+\tan^2(\alpha))\sin(µ\alpha)-(8\tan(\alpha)+3)\cos(\alpha)}{\sin^2(\alpha)}$$
A: 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:
$$\tan \alpha = \frac 3x \Rightarrow \alpha = \arctan \frac{3}{2\sqrt[3]{9}}=0.624763... \ \text{rad}=35.8^\circ.\\
L(2\sqrt[3]{9})=14.99216...$$
A: Using SymPy (and $t$ instead of $\alpha$) to compute the first derivative of $L$,
>>> from sympy import *
>>> t = Symbol('t') 
>>> L = (8 + (3/tan(t))) / (cos(t))
>>> simplify(diff(L,t))
8*sin(t)/cos(t)**2 - 3*cos(t)/sin(t)**2

Hence, finding where the first derivative of $L$ vanishes produces the equation
$$8 \,\frac{\sin{\left(t \right)}}{\cos^{2}{\left(t \right)}} = 3 \,\frac{ \cos{\left(t \right)}}{\sin^{2}{\left(t \right)}}$$
or, assuming that neither denominator is zero,
$$\left( \frac{\sin {\left(t \right)}}{\cos{\left(t \right)}} \right)^3 = \frac 38$$
Computing floating-point values of the arctangent of $\sqrt[3]{\frac 38}$ and the corresponding value of $L$,
>>> atan(cbrt(Rational(3,8))).evalf()
0.624763427949745
>>> L.subs(t,0.624763427949745)
14.9921648882861

