Question 1048773
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If the sum of the height and the radius is 60, then the height as a function of the radius is *[tex \Large 60\ -\ r].  So the volume as a function of the radius is then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ V(r)\ =\ \pi r^2(60\ -\ r)\ =\ \pi(60r^2\ -\ r^3)]


Then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dV}{dr}\ =\ \pi(120r\ -\ 3r^2)]


Which has a zero at *[tex \Large r\ =\ 40]


Since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2V}{dr^2}\ =\ 120\ -\ 6r]


equals *[tex \Large -120] when *[tex \Large r\ =\ 40], *[tex \Large r\ =\ 40] is a local maximum of the original function.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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