Simplifying Cubic: Find Local Extrema

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Homework Help Overview

The discussion revolves around finding the local extrema of a cubic function derived from a given expression involving a radius variable, r. The original poster presents their simplification process and expresses uncertainty about further simplifying the resulting cubic equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the simplification of a cubic equation derived from a derivative set to zero. There are mentions of using the rational root theorem and questioning the approach to finding roots of the cubic.

Discussion Status

The discussion is ongoing, with participants exploring different methods for simplifying the cubic equation. Some guidance has been offered regarding the rational root theorem, but there is no explicit consensus on the best approach yet.

Contextual Notes

There are constraints regarding the radius variable, which cannot be zero. Participants also note confusion around the simplification process and the application of the rational root theorem with integer constants.

ciubba
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I need to find the local extrema of

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

which I derived and simplified to

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

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

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

I used the binomial theorem and more algebra to obtain

[tex]r^3+1.5r^2+.75r-7.875[/tex]

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

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

which I derived and simplified to

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

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

I used the binomial theorem and more algebra to obtain

[tex]r^3+1.5r^2+.75r-7.875[/tex]

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

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

which I derived and simplified to

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

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

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

I used the binomial theorem and more algebra to obtain

[tex]r^3+1.5r^2+.75r-7.875[/tex]

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.
 

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