The question:
I'm given the figure shown above, and need to calculate the length of the longest rod that can fit inside this figure and rotate the corner.
My thoughts:
I have tried doing the following : put $(0,0)$ at the bottom left corner. This way, the place where the rod touches the upper block is $(2,1) $ , and if we denote by $(2+t,0)$ the place where the rod touches the lower block, we get that the place where it touches the lower block is $y=t+2 $ , and then, the length is $d=\sqrt{2}(t+2)$ which doesn't have a maximum.
What am I doing wrong ?
THe final answer should be $ \sqrt{(1+\sqrt[3]{4} ) ^2 + (2+\sqrt[3]{2})^2 } $ .
Thanks !
$w,h,r$ above with 17 rectangles within them">
Best Answer
$y = 2+\frac{2}{x-1}$
$ L = \sqrt{x^2 + (2+\frac{2}{x-1})^2}$
Take the derivative of the function inside the square root and equate it to 0
$\frac{dL}{dx} = (x-1)^{3} - 4 = 0$
$x=4^{1/3} + 1$
Thus $L = \sqrt{(4^{1/3} + 1)^2 + (2+2^{1/3})^2}$