Optimizing trigonometric equation I've come across a problem from an old calculus textbook which goes like
A tool shed, $250\space cm$ high and $100\space cm$ deep is build against a wall. Calculate the shortest ladder length that can reach from the ground to the wall behind.
I got 
$\text{length} = \frac{100}{cosx} + \frac{250}{sinx}$
Which I derived to get 
$\text{Length'}=\frac{100sinx}{cos^2(x)} - \frac{250cosx}{sin^2(x)}$
Set that to equal 0 and got 
$250\cos^3x=100\sin^3x$
And now stuck would like to know how to finish this problem!
 A: Since you have the answer, let me suppose that you need to compute $x$ and that you only have a very limited calculator (no trigonometric function available).
You obtained $\tan^3(x)=2.5$ which means $\tan(x)\approx 1.35721$ which means that $x$ is slightly larger than $\frac \pi 4$. So, consider Taylor expansion built at $x=\frac \pi 4$ $$\tan(x)=1+2 \left(x-\frac{\pi }{4}\right)+2 \left(x-\frac{\pi
   }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ So, ignoring higher terms, you need to solve $$1+2y+2y^2=\sqrt[3]{\frac{5}{2}}$$ where $y=x-\frac{\pi }{4}$. This gives $$y=\frac{1}{2} \left(\sqrt{2^{2/3} \sqrt[3]{5}-1}-1\right)\approx 0.154679$$ that is to say $x \approx 0.940077$ while the exact solution is $\approx 0.935793$ which is not too bad for the price of a quadratic.
Another way, may be simpler could be to consider that, from the series $$\tan^3(x)=1+6 \left(x-\frac{\pi }{4}\right)+18 \left(x-\frac{\pi
   }{4}\right)^2+O\left(\left(x-\frac{\pi }{4}\right)^3\right)$$ which leads to the equation $$1+6y+18y^2=\frac 52$$ from which $y=\frac 16$ and $x=\frac{\pi }{4}+\frac 16\approx 0.952065$. The result is worse in quality but nicer at least.
