Simplifying Cubic: Find Local Extrema

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I need to find the local extrema of

\pi r^2(\frac{16}{(r+.5)^2}-1)

which I derived and simplified to

\frac{16 \pi r}{(r+.5)^3}=2 \pi r

which simplifies to \frac {16 \pi r}{2 \pi r}=(r+.5)^3

The radius cannot be zero, so I simplified 8=(r+.5)^3

I used the binomial theorem and more algebra to obtain

r^3+1.5r^2+.75r-7.875

Now I am unsure of how to simplify the cubic. Normally I would use rational roots, but I don't know how to do that with an integer constant. I need either a method of simplifying this cubic or a place where I could have simplified the derivative better.
 
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8 = (r+0.5)3 , why not to take 3rd root ?
 
Oh my god, I can't believe I missed that.
Thanks, I guess.
 
ciubba said:
I need to find the local extrema of

\pi r^2(\frac{16}{(r+.5)^2}-1)

which I derived and simplified to

\frac{16 \pi r}{(r+.5)^3}=2 \pi r
... which you differentiated, simplified, and then set to zero.
ciubba said:
which simplifies to \frac {16 \pi r}{2 \pi r}=(r+.5)^3

The radius cannot be zero, so I simplified 8=(r+.5)^3

I used the binomial theorem and more algebra to obtain

r^3+1.5r^2+.75r-7.875

Now I am unsure of how to simplify the cubic. Normally I would use rational roots, but I don't know how to do that with an integer constant. I need either a method of simplifying this cubic or a place where I could have simplified the derivative better.
 
ciuba,ciuba you messed it up more than once
 
ciubba said:
I need to find the local extrema of

\pi r^2(\frac{16}{(r+.5)^2}-1)

which I derived and simplified to

\frac{16 \pi r}{(r+.5)^3}=2 \pi r

which simplifies to \frac {16 \pi r}{2 \pi r}=(r+.5)^3

The radius cannot be zero, so I simplified 8=(r+.5)^3

I used the binomial theorem and more algebra to obtain

r^3+1.5r^2+.75r-7.875

Now I am unsure of how to simplify the cubic. Normally I would use rational roots, but I don't know how to do that with an integer constant. I need either a method of simplifying this cubic or a place where I could have simplified the derivative better.

Try the full-fledged "rational root theorem' as described in http://en.wikipedia.org/wiki/Rational_root_theorem --- it works. However, you need to convert to integer coefficients throughout.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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