Find the maximimum value of $L$

Question

Find the maximum length of rod $L$ in terms of $a$ and $b$ that it can passes such channel. Note : Suppose the problem in two dimension and pass up the thickness of rod.

Answer 1

So from above graph

$\displaystyle \sin \theta = \frac{a}{l_{1}}\Rightarrow l_{1}=\frac{a}{\sin \theta}$

and $\displaystyle \cos \theta = \frac{b}{l_{2}}\Rightarrow l_{2}=\frac{b}{\cos \theta}$

So $$L = l_{1}+l_{2} = \frac{a}{\sin \theta}+\frac{b}{\cos \theta}\;, \theta \in \left(0,\frac{\pi}{2}\right)$$

Now Use Holder,s Inequality

$$\left(\sin^2 \theta +\cos^2 \theta \right)\left(\frac{a}{\sin \theta}+\frac{b}{\cos \theta}\right)\left(\frac{a}{\sin \theta}+\frac{b}{\cos \theta}\right)\geq \bigg[a^{\frac{2}{3}}+b^{\frac{2}{3}}\bigg]^3$$

So $$\left(\frac{a}{\sin \theta}+\frac{b}{\cos \theta}\right)\geq \bigg(a^{\frac{2}{3}}+b^{\frac{2}{3}}\bigg)^{\frac{3}{2}}$$

and equality hold when $\displaystyle \frac{\sin^3 \theta}{a}=\frac{\cos^3 \theta}{b}\Rightarrow \tan \theta = \left(\frac{a}{b}\right)^{\frac{1}{3}}$

Answer 2

let $\alpha$ the angle in the right triangle build with $l$ then we have $$\sin(\alpha)=\frac{a}{x}$$ and $$\cos(\alpha)=\frac{b}{l-x}$$ thus we have found $$l=\frac{a}{\sin(\alpha)}+\frac{b}{\cos(\alpha)}$$ with $$0<\alpha<\frac{\pi}{2}$$ no you can differentiate $$l=l(\alpha)$$ with respect to $$\alpha$$