Finding the value of r for which A is a minimum

[kais]

Hey guys im stuck on a question, any help would be appreciated. Question goes

find the value of r for which A is a minimum, and calculate the minimum area of a container

A=pi r^2 + 1000/r

now ive got the derivative which is 2pi r - 1000/r^2

this is where i had trouble finding the zeroes of the derivative, i got the equation to equal

r^3=500/pi

im not sure what i have to do to the equation to turn r^3 into r? or are there easier ways to get the zeroes in this equation?

cheers in advanced

 

[stapel]

Find the value of r for which A is a minimum, and calculate the minimum area of a container

A=pi r^2 + 1000/r

I'm going to assume, based on what follows, that the above means the following:

. . . . .$\displaystyle A\, =\, \pi\, r^2\, +\, \dfrac{1000}{r}$

(In other words, and in contrast to many posters, you have formatted the equation correctly.)

Now I've got the derivative which is 2pi r - 1000/r^2

In other words (and more correctly):

. . . . .$\displaystyle \dfrac{dA}{dr}\, =\, 2\, \pi\, r\, -\, \dfrac{1000}{r^2}$

This is where I had trouble: finding the zeroes of the derivative. I got the equation to equal

r^3=500/pi

I'm not sure what I have to do to the equation to turn r^3 into r?

This is just algebra, so think back to what you did back in algebra:

When you had "(this squared) equals (that number)", you solved by taking square roots (and adding a "plus, minus" to the numerical side). Now, you have "(this cubed) equals (that number)", you can solve by taking [what kind of] roots (with no "plus, minus" necessary).