Finding the zero of a weird cubic function Volume of Silo: A silo grain consists of a cylindrical main section and a hemispherical roof. If the total volume of the silo (including the part inside the roof section) is $\mathbf{15,000\ \ ft^3}$ and the cylindrical part is $\mathbf{30\ ft}$ tall, what is the radius of the silo, correct to the nearest tenth of a foot?
I already got to the equation
$$r^3+45r^2-\frac{22500}{\pi} = 0$$
but I don't know how to solve it without a graphing calculator. (The answer with the calculator is $\mathbf{11.28\ ft}$, knowing that the radius should be positive.)
How do I solve this? (I can use some calculus including limits and derivatives.)
This is a reference for the question I am trying to solve.
 A: You face the problem of solving a cubic polynomial equation in $r$ $$r^3+45r^2-\frac{22500}{\pi} = 0$$ For sure, you could use Cardano method and show that there are three real roots, only one being positive. The analytical formula will be quite ugly.
The other solution is to use numerical method for finding the zero of $$f(r)=r^3+45r^2-\frac{22500}{\pi}$$ and Newton method will probably be the simplest. As you probably remember, starting from a "reasonable" guess $r_0$, Newton method will update it according to $$r_{n+1}=r_n-\frac{f(r_n)}{f'(r_n)}$$ For this case, the iterates can write (after some minor simplifications $$r_{n+1}=\frac{2 \pi  r_n^3+45 \pi  r_n^2+22500}{3 \pi  r_n^2+90 \pi  r_n}$$
Looking a the plot of the function, you notice that there is a solution close to $r=10$; so, let us use $r_0=10$; the successive iterates will then be $$x_1=11.38497703$$ $$x_2=11.28132043$$ $$x_3=11.28071248$$ $$x_4=11.28071246$$ which is the solution for ten significant figures.
Edit
Just for your curiosity (do not worry, you will learn about that later),we can also approximate, around a given point, functions by Padé approximants. In this case, using the simplest, we should get (around $r=10$) $$r^3+45r^2-\frac{22500}{\pi}\approx \frac{\frac{25 (225+137 \pi ) (r-10)}{4 \pi }+\frac{500 (11 \pi -45)}{\pi
   }}{\frac{10-r}{16}+1}$$ and setting the numerator to zero would give as an approximation $$r=\frac{5850+490 \pi }{225+137 \pi }\approx 11.27464261$$ Doing the same around $r=11$ would give $$r=\frac{16582500+1718321 \pi }{585000+434027 \pi }\approx 11.28065596$$
