Re: Numerical method help

Can you give an idea of how you are supposed to use numerical methods for this?

The general method is to equate the derivative of the surface area with respect to the radius to zero and solve for the radius.

When you do this you get a closed form solution for r that can be evaluated on any calculator.

Is there some other method your teacher has in mind?

Re: Numerical method help

We have to use excel for this and create a graph after. The radius is a fixed value. I don't know what to do

Re: Numerical method help

Quote:

Originally Posted by

**RandomUsername** We have to use excel for this and create a graph after. The radius is a fixed value. I don't know what to do

Well to do it numerically you're going to have to pick a number for the volume.

Then I'd put a list of radii in column A and plot the surface area according to your formula in column B.

Chart that. You'll be easily able to pick out the value of r in your list that makes the smallest surface area.

Without a better idea of what methods you have available to use I'm not sure how to suggest you go about finding the actual minimum.

Re: Numerical method help

The volume is 300ml. so what i do is do the formula and chart it?

Re: Numerical method help

If you are given that the volume is 300 ml (your original post did not say that the volume was constant) then, yes, you can use the formula you give, $\displaystyle A= 2\pi r^2+ 2\frac{V}{r}= 2\pi r^2+ \frac{600}{r}$.

Yes, you could graph that and "zoom" in on points were the area is maximum. Another way to do this is to differentiate the function, with respect to r, to get $\displaystyle 4\pi r- \frac{600}{r^2}$. Set that equal to 0 and solve $\displaystyle 4\pi r^3= 600$. I don't think you really need a numerical method to solve that.

Re: Numerical method help

Quote:

Originally Posted by

**HallsofIvy** If you are given that the volume is 300 ml (your original post did not say that the volume was constant) then, yes, you can use the formula you give, $\displaystyle A= 2\pi r^2+ 2\frac{V}{r}= 2\pi r^2+ \frac{600}{r}$.

Yes, you could graph that and "zoom" in on points were the area is maximum. Another way to do this is to differentiate the function, with respect to r, to get $\displaystyle 4\pi r- \frac{600}{r^2}$. Set that equal to 0 and solve $\displaystyle 4\pi r^3= 600$. I don't think you really need a numerical method to solve that.

Thank you very much. This has really helped