# Two halls 6 and 9 meters perpendicularly intersect. Optimization

Two halls 6 and 9 meters perpendicularly intersect. Find the length of the longest straight bar to be passed horizontally from one aisle to another by a corner without deformation.

and this is my try:

How to find the equation to maximize in this problem.Please.

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I can't tell what you are doing in the computation. It looks like you have the right idea-to find the shortest segment that goes from one outer wall to the other, touching the inner corner. Please explain your reasoning for each line. Preferably, format it with $\LaTeX$. A tutorial is here – Ross Millikan May 7 '14 at 20:41
Am I missing something, or is the answer basically infinity? You can just keep making it steeper forever and the length will keep increasing. – hacatu May 8 '14 at 0:15
@user2171981 I guess you should be able to carry it around the corner. – Martin Ueding May 8 '14 at 16:38

We use your diagram. We calculate the length of the diagonal when the angle at the bottom is $\theta$.

The part from the bottom to the obstructive corner is $\frac{9}{\sin \theta}$, and the rest is $\frac{6}{\cos\theta}$. We want to minimize $f(\theta)$, where $$f(\theta)=\frac{9}{\sin\theta}+\frac{6}{\cos\theta}.$$ To find the minimum, use the usual tools. We have $$f'(\theta)=-\frac{9\cos\theta}{\sin^2\theta}+\frac{6\sin\theta}{\cos^2\theta}.$$ Set this equal to $0$. You will find that $\tan^3\theta$ has to be a certain quantity.

Remark: A rectangular coordinates approach along the lines you are pursuing will also work, albeit a little less smoothly. I cannot make detailed comments, the work is difficult to read.

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and to find the maximum its the same ? well fund the f´´where is negative and that is the maximum – Knight May 7 '14 at 20:59
There is no maximum of $f(\theta)$. The largest rod that can be carried is the minimum of $f(\theta)$. – André Nicolas May 7 '14 at 21:01
so that´s why i didn't found the max, really ty!!!ç – Knight May 7 '14 at 21:03
So is it better to do this with or without trigonometric functions? (My posted answer here does not use them.) – Michael Hardy May 7 '14 at 21:04
I should have drawn a labelled diagram. Let $B$ be the bottom of the rod (in the picture) and let $C$ be the corner where the rod meets the two walls. Let $A$ be the point on the bottom wall below $C$. Then $\theta=\angle CBA$. And $\frac{9}{\sin\theta}$ is the length of $BC$. This is because $AC=9$. The other term is the length of the rest of the rod. – André Nicolas May 8 '14 at 2:58

This superficially appears to be a maximization problem but is really a minimization problem. You've drawn some diagonal lines through the corner that you seem to have labeled $(-6,9)$. You have to figure out which value of the coordinate you've called $x$ makes that diagonal line as short as possible.

You've got \begin{align} g(x) = (x+6)^2 + \left( 9 + \frac{54}{x} \right)^2 & = (x+6)^2 + 81\left(\frac{x+6}{x}\right)^2 \\[10pt] & = (x+6)^2\left( 1 + \frac{81}{x^2} \right). \end{align} You need the value of $x$ that minimizes that.

Notice that $g(x)\to\infty$ as $x\downarrow0$ and $g(x)\to\infty$ as $x\to\infty$, and the function is continuous on $(0,\infty)$, so it must have a global minimum somewhere in between. If there's only one place in that interval where $g'=0$, then that must be it.

Alright, in response to comments: \begin{align} g'(x) & = (x+6)^2 \frac{d}{dx}\left( 1 + \frac{81}{x^2} \right) + \left( 1 + \frac{81}{x^2} \right) \frac{d}{dx}(x+6)^2 \\[10pt] & = (x+6)^2 \cdot\frac{-162}{x^3} + \left( 1 + \frac{81}{x^2} \right)2(x+6). \end{align}

This is $0$ when $x=-6$, and at other points we can divide by $x+6$ both sides of the equation that sets the expression above equal to $0$. We get $$(x+6) \cdot\frac{-162}{x^3} + \left( 1 + \frac{81}{x^2} \right)2 = 0.$$ Multiplying both sides by $x^3$ we get $$(x+6)(-162) + (x^3 + 81x)2 = 0$$ $$2x^3 - 972 = 0 ,$$ $$x^3 - 486 = 0$$ $$x = \sqrt[3]{486} = 3\sqrt[3]{18} \approx 7.86.$$

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i did that but when i try to find the max, but the values i got doesn't match with the graph – Knight May 7 '14 at 21:06
i suppose that i need more condition or something like that but idk how to solve this, the other guy told me that there is no max – Knight May 7 '14 at 21:06
There is no max, but as has been emphasized by Michael Hardy, we are looking for the minimum length of the diagonal. Max length is the minimum of the diagonal, the shortest diagonal is what's stopping us from carrying the thing around the corner. Unless the hall happens to have non-zero height! – André Nicolas May 7 '14 at 21:14
okk...... i understand now because i construct my equation as Michael but i stuck when try to find max, but ty for explain me. – Knight May 7 '14 at 21:18
@Knight : I've added a bit on finding the point where the derivative is $0$. – Michael Hardy May 7 '14 at 21:40