The problem is question 3 of what I am about to download into this question. I drew a diagram of what the problem actually is, my professor has verified it's correct. I don't want an exact answer to this question, just a way to convert question 3 into an equation so that I can use programming to solve. I have software I have created that can solve for zeros of functions and minimums of functions numerically.

I am thinking that this is a problem of looking for some type of maximum because we want the biggest ladder that can fit through. Is this correct?

It's probably a nooby question, but my trignometry is pretty rusty. I have a feeling it will be blatantly obviousenter image description here


marked as duplicate by David K, Jonas Meyer, user147263, user99914, Prasun Biswas May 20 '15 at 3:59

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  • $\begingroup$ The "duplicate of" question is slightly different, but I think you'll find it has some good ideas about how to make your diagram into formulas. Just set $A=7$ and $B=9$. $\endgroup$ – David K May 20 '15 at 0:35

Here's how I think of it. You can break it into two triangles, one with base $7$, hypotenuse $L_1$ and angle $\theta \neq 0$ and a second triangle of height $9$, hypotenuse $L_2$ and identical angle $\theta$. (Depending on where you put theta) you get the equations $$\cos(\theta)=\frac{7}{L_1}$$ and $$\sin(\theta)=\frac{9}{L_2}$$ use these two equations to get $L = L_1+L_2$. You can minimize $L$ and get your answer.

  • $\begingroup$ @DavidK thanks for catching that, you are indeed correct $\endgroup$ – graydad May 20 '15 at 2:21
  • $\begingroup$ This is about as simple as the explanation can get. And it works. $\endgroup$ – David K May 20 '15 at 2:22
  • $\begingroup$ Is there anyway to translate this problem into a function of one variable in which a minimum of this function's graph is the answer? $\endgroup$ – user121615 May 21 '15 at 11:43
  • $\begingroup$ This is a function of one variable, just $\theta$ $\endgroup$ – graydad May 21 '15 at 13:31

So you have: $l$ is the length of the ladder, and $\theta$ is the angle between the ladder and the wall atop the T of the junction.

Find the minimum of $\;l = 9\csc \theta + 7\sec \theta\;$ for $\;\theta\in (0;\pi/2)\;$ .   This will be the longest ladder which can navigate the corner.   (Any longer and it cannot pass the narrowest part of the turn.)

$$\begin{align} l & =\min_{\theta\in(0;\pi/2)}\left( \frac 9 {\sin\theta} + \frac 7 {\cos\theta}\right) \\[2ex] & \boxed{\begin{align} 0 & = \frac{-9\cos\theta}{\sin^2\theta}+\frac{7\sin\theta}{\cos^2\theta} \\[1ex] 0 & = -9\cos^3 \theta + 7 \sin^3 \theta \\[1ex] \tan^3 \theta & = 9/7 \\[1ex] \theta & = \arctan\sqrt[3]{\tfrac 97} \end{align}} \\[2ex] l & = \frac 9{\sin\arctan\sqrt[3]{\tfrac 97}}+\frac 7{\cos\arctan\sqrt[3]{\tfrac 97}} \\[1ex] & = (3 \sqrt[3]{21}+7) \sqrt{1+3 \sqrt[3]{\tfrac 3{7^2}}} \end{align}$$

  • $\begingroup$ Is there anyway to turn this question into a 'function', so to speak? So that I can plot it on a graph and see where this minimum lies? $\endgroup$ – user121615 May 21 '15 at 11:22
  • $\begingroup$ @user121615 Plot $y= 9/\sin(x) + 7/\cos(x)$ $\endgroup$ – Graham Kemp May 21 '15 at 13:30

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