# The fireman's problem (optimization)

• October 16th 2008, 04:58 PM
elitespart
The fireman's problem (optimization)
A fence a feet high is b feet from a high burning building. Find the length of the shortest ladder that will reach from the ground across the top of the fence to the building.

I have the picture set up and now I'm stuck. I know that I have to express the ladder in terms of x and I'm pretty sure that it involves similar triangles. Thanks.
• October 16th 2008, 07:49 PM
Soroban
Hello, elitespart!

I found that Trig made this easier . . .

Quote:

A fence $a$ feet high is $b$ feet from a burning building.
Find the length of the shortest ladder that will reach from the ground
across the top of the fence to the building.

Code:

                              * R                           *  |                         *    |                     * θ      |                 Q * - - - - - * S               *  |    b    |             *    |a          |         * θ      |          |       * - - - - - * - - - - - *       P          U    b    T

The fence is $QU = a$ feet.
It is $UT = b$ feet from the building.
. . Note that $QS = b$.

The ladder is $PR = L$, which makes angle $\theta$ with the ground.
.Note that $\angle RQS = \theta$.

In right triangle $QUP\!:\;\;\csc\theta = \frac{PQ}{a}\quad\Rightarrow\quad PQ = a\csc\theta$

In right triangle $RSQ\!:\;\;\sec\theta = \frac{QR}{b} \quad\Rightarrow\quad QR = b\sec\theta$

We have: . $PR \;=\;L \;=\;a\csc\theta + b\sec\theta$ .[1]

Minimize $L$, solve $\frac{dL}{d\theta} = 0$

. . $\frac{dL}{d\theta} \;=\;-a\csc\theta\cot\theta + b\sec\theta\tan\theta \;=\;0$

We have: . $-a\cdot\frac{1}{\sin\theta}\cdot\frac{\cos\theta}{\ sin\theta} + b\cdot\frac{1}{\cos\theta}\cdot\frac{\sin\theta}{\ cos\theta} \;=\;0$ . . $\Rightarrow\quad \frac{b\sin\theta}{\cos^2\!\theta} - \frac{a\cos\theta}{\sin^2\!\theta} \;=\;0$

Multiply by $\sin^2\!\theta\cos^2\!\theta\!:\;\;b\sin^3\!\theta - a\cos^3\!\theta \;=\;0 \quad\Rightarrow\quad b\sin^3\!\theta \;=\;a\cos^3\!\theta$

. . $\frac{\sin^3\!\theta}{\cos^3\!\theta} \;=\;\frac{a}{b} \quad\Rightarrow\quad \left(\frac{\sin\theta}{\cos\theta}\right)^3 \;=\;\frac{a}{b}$ . . $\Rightarrow\quad \tan^3\!\theta \;=\;\frac{a}{b} \quad\Rightarrow\quad \tan \theta \;=\;\sqrt[3]{\frac{a}{b}}$

We have: . $\tan\theta \;=\;\frac{\sqrt[3]{a}}{\sqrt[3]{b}} \;=\;\frac{opp}{adj}$

Using Pythagorus, we have: . $\begin{Bmatrix}opp &=& a^{\frac{1}{3}} \\ adj &=& b^{\frac{1}{3}} \\ hyp &=& \sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}} } \end{Bmatrix}$

Then: . $\csc\theta \;=\;\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}} {a^{\frac{1}{3}}} \qquad \sec\theta \;=\;\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}}{b^{\frac{1}{3}}}$

Substitute into [1]: . $L \;=\;a\cdot\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}}{a^{\frac{1}{3}}} + b\cdot\frac{\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}} {b^{\frac{1}{3}}}$ . $= \;a^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}} + b^{\frac{2}{3}}\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}$

Factor: . $L \;=\;\sqrt{a^{\frac{2}{3}} + b^{\frac{2}{3}}}\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)$

Therefore: . $\boxed{L \;=\;\left(a^{\frac{2}{3}} + b^{\frac{2}{3}}\right)^{\frac{3}{2}}}$

• October 16th 2008, 08:52 PM
elitespart
Wow! Thank you so much. I really appreciate it.
• March 25th 2012, 06:46 PM
ftmtmr
Re: The fireman's problem (optimization)
i had the same question in my study guide :/ i couldn't get the answer :S can u explain more deeply?