2.2 Example

From [1, p. 116, 30]

$$\epsffile{optx11.eps}$$

Given: A free standing wall, located $3 \frac38$ ft from the side of a house.

Asked: What is the length $\ell$ of the shortest ladder that can reach the house (over the free standing wall).

2.2.1 Definition

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Two degrees of freedom: say $h$ and $d$

One inequality constraint: the ladder must be above the free standing wall.

2.2.2 Reduction

The shortest ladder hits the free standing wall:

$$\epsffile{optx12.eps}$$

One degree of freedom left: $\varphi$.

2.2.3 Further reduction

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At the minimum:

$$\frac{{\rm d}\ell}{{\rm d}\varphi} = 0$$

2.2.4 Finding the length

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First find $a$:

$$a = \frac{8}{\tan\varphi}.$$

Then:

$$\ell = \frac{{\textstyle 3\frac38}+a}{\cos\varphi} = \frac{\textstyle 3\frac38}{\cos\varphi} + \frac{8}{\sin\varphi}$$ (2.1)

2.2.5 Finding the optimum angle

$$\frac{{\rm d}\ell}{{\rm d}\varphi} = \frac{\textstyle 3\... ...phi} \sin\varphi - \frac{8}{\sin^2\varphi} \cos\varphi = 0.$$

$$\frac{27}{8\cos^2\varphi} \sin\varphi = \frac{8}{\sin^2\varphi} \cos\varphi$$

$$\tan^3 \varphi = \frac{64}{27} \quad\quad\Rightarrow\quad\quad \varphi_{\mbox{min}} = 0.9273 \mbox{ radians }$$

2.2.6 Finding the optimum length

From (2.1)

$$\ell_{\mbox{min}} = 15.625 \mbox{ ft }$$