# Maximum Length of a Ladder Problem

**Problem:** Find the maximum length, *L*_{max}, of a ladder that can pass through a hall of width of *w*_{1} to a hall of width *w*_{2} if the ladder is carried exactly parallel to the floor.

Let us consider the length, *L*, of the ladder as a function of *θ*, the angle which it makes with respect to one of the walls of the hall as shown in Figure 1. We have the following equalities from the figure:

(1) $\cos\theta = \frac{w_1}{x}$

(2) $\sin\theta = \frac{w_2}{L-x}$

Solving for *x* in (i) and substitution into (ii) gives:

(i) $L = \frac{w_1}{\cos\theta} + \frac{w_2}{\sin\theta}$

To find maximum of *L*, we take the derivative of *L* with respect to *θ* and set the derivative to 0 to solve the value of *θ* that will maximize *L*.

(ii) $\frac{dL}{d\theta} = w_1\frac{\sin\theta}{\cos^2\theta} - w_2\frac{\cos\theta}{\sin^2\theta} = 0$

(iii) $w_1\sin^3\theta - w_2\cos^3\theta = 0$

(iv) $\tan^3\theta = \frac{w_2}{w_1}$

Therefore, the angle that maximizes *L* is $\theta = \arctan\sqrt[3]{\frac{w_2}{w_1}}$. To find the maximum length of the ladder, we substitute this value into (i).

(v) $L = \frac{w_1}{\cos\left(\arctan\sqrt[3]{\frac{w_2}{w_1}}\right)} + \frac{w_2}{\sin\left(\arctan\sqrt[3]{\frac{w_2}{w_1}}\right)}$

We can use a right triangle with sides of $w_1^{\frac{1}{3}}$ and $w_2^{\frac{1}{3}}$ to find the sine and cosine of this angle.

(vi) $\sin\left(\arctan\sqrt[3]{\frac{w_2}{w_1}}\right) = \frac{w_2^{\frac{1}{3}}}{\left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}$

(vii) $\cos\left(\arctan\sqrt[3]{\frac{w_2}{w_1}}\right) = \frac{w_1^{\frac{1}{3}}}{\left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}$

Substituting (vi) and (vii) into (i), we get:

(viii) $L = \frac{w_1}{\frac{w_1^{\frac{1}{3}}}{\left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}}} + \frac{w_2}{\frac{w_2^{\frac{1}{3}}}{\left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}}$

(ix) $L = \left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}\cdot w_1^{\frac{2}{3}} + \left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}\cdot w_2^{\frac{2}{3}$

(x) $L = \left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{1/2}\cdot\left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)$

(x) $L = \left(w_1^{\frac{2}{3}} + w_2^{\frac{2}{3}}\right)^{3/2}$

Therefore, our maximum length of the ladder is $L_{\text{max}} = \left( w_1^{2/3} + w_2^{2/3} \right)^{3/2}$.