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\[\large 0=\frac{ r-2700 }{ r^2 }+62 \pi r\]
You are trying to find the area of a radius?
Sorry I meant the area of a circle.

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No it's an optimization problem I am trying to do, I just need to solve for 'r'
Oh I thought that because the formula to finding the area of a circle,you have to first square the radius and then multiply it by pi (r^2)3.14.
Multiply everything by r^2. Then factor out an r. From there it's a quadratic.
Oh my bad. You actually get a cubic.
top and bottom or just top?
Multiply everything by r^2. When you multiply the fraction, multiply on top.
\[r^2 or \frac{ r^2 }{ r^2 }\]
Just r^2
yeah so woulnt that cancel out with the fraction
Yes =) That's the point. It gets r out of the denominator.
ohh i thought you were also telling me to muktiply the top by r^3 so the function would be diff, i misunderstood
\(\huge 0 = \frac{r-2700}{r^2} + 62\pi*r\) multiply everything by r^2 \(\huge (0)*r^2 = (\frac{r-2700}{r^2})*r^2 + (62\pi*r)*r^2\)
Gives: \(\huge 0 = r-2700 + 62\pi*r^3\)
And honestly from there your best bet is to use a cubic solver of some sort. Wolfram Alpha should do just fine.
|dw:1371520632051:dw| yeah but the thing is like during an exam i cant use it :P
Are you sure the question is correct?
I suggest you double check, because you will not solve this without the aid of wolfaplha
here is the question and my solution
1 Attachment
I suggest you take a look at this http://gbbservices.com/math/cubic.html
Okay on the step where you take C', you make incorrect use of the quotient rule. The derivative of the top will be 0.
The correct derivative is: \(\Large C' = \frac{-27900}{r^2} + 62\pi*r\)
Set the derivative equal to 0 and solve: \(\Large \frac{-27900}{r^2} + 62\pi*r = 0\) \(\Large (\frac{-27900}{r^2} + 62\pi*r)*r^2 = 0*r^2\) \(\Large -27900 + 62\pi*r^3 = 0\) \(\Large 62\pi*r^3 = 27900\) You good from there?

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