Two halls 6 and 9 meters perpendicularly intersect. Optimization

Question

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.

Answer 1

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.

Answer 2

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 $$ (x+6)(-162) + (x^3 + 81x)2 = 0 $$ $$ 2x^3 - 972 = 0 , $$ $$ x^3 - 486 = 0 $$ $$ x = \sqrt[3]{486} = 3\sqrt[3]{18} \approx 7.86. $$