Optimization Problem 2

[shouav1]

It may look weird, but this (π) is pi.

Building a cylinder the volume of 2750π with the following costs:

Top: $30(area of top)= 30(πr²)
Sides: $48(area of sides)=48(2πrh)
Bottom: $50(area of bottom)= 50(πr²)
Welding: $30(Circumference of top and bottom)=30(2(2πr))

Cost function: C(r)=30πr²+50πr²+96πrh+120πr
V=2750π=πr²h
h=(2750π)/(πr²)=2750/r²

plug in h
C(r)= 30πr²+50πr²+96πr(2750/r²)+120πr

simplify
C(r)=80πr²+(264000π)/r+120πr

differentiate
C’(r)=160π-(264000π/r²)+120π

Now I am suppose to simplify the derivative into a cubic polynomial before applying newton's method to get the correct roots, but I am not really sure how to.
I got (160πr³-264000π+120πr²)/r² but I am not sure if it's correct. Can I simplify it any other way to get a cubic polynomial?

 

[Deleted member 4993]

It may look weird, but this (π) is pi.

Building a cylinder the volume of 2750π with the following costs:

Top: $30(area of top)= 30(πr²)
Sides: $48(area of sides)=48(2πrh)
Bottom: $50(area of bottom)= 50(πr²)
Welding: $30(Circumference of top and bottom)=30(2(2πr))

Cost function: C(r)=30πr²+50πr²+96πrh+120πr
V=2750π=πr²h
h=(2750π)/(πr²)=2750/r²

plug in h
C(r)= 30πr²+50πr²+96πr(2750/r²)+120πr

simplify
C(r)=80πr²+(264000π)/r+120πr

differentiate
C’(r)=160πr-(264000π/r²)+120π

Now I am suppose to simplify the derivative into a cubic polynomial before applying newton's method to get the correct roots, but I am not really sure how to.
I got (160πr³-264000π+120πr²)/r² but I am not sure if it's correct. Can I simplify it any other way to get a cubic polynomial?

C'(r) looks correct to me. However, it is not a polynomial.

You do not need to have a polynomial to apply Newton's method.

 

[shouav1]

The problem is that I can't get the correct root with the current derivative. Applying Newton's method always gets me at -15.04....
I need to get the other root at around 11.57...

 

[MarkFL]

We have:

\(\displaystyle C'(r)=160\pi r-\dfrac{264000\pi}{r^2}+120\pi=0\)

Multiply through by \(\displaystyle \dfrac{r^2}{40\pi}\ne0\):

\(\displaystyle f(r)=4r^3+3r^2-6600=0\)

\(\displaystyle f'(r)=12r^2+6r\)

Newton's method then gives us:

\(\displaystyle r_{n+1}=r_n-\dfrac{4r_n^3+3r_n^2-6600}{12r_n^2+6r_n}\)

Looking at the graph of \(\displaystyle f(r)\) we see that \(\displaystyle r\approx12\) is a reasonable initial guess for the 1 real root. So, we have:

\(\displaystyle r_0=12\)

\(\displaystyle r_1=11.58\bar{6}\)

\(\displaystyle r_2\approx11.5718905731\)

\(\displaystyle r_3\approx11.5718720809\)

\(\displaystyle r_4\approx11.5718720809\)

I have reached the limit of accuracy of my calculator. I have a TI-89, and the method I used to apply Newton's method here is to enter:

12 [ENTER]

ans(1)-(4ans(1)^3+3ans(1)^2-6600)/(12ans(1)^2+6ans(1)) [♦][ENTER]

Then keep pushing [♦][ENTER] for successive approximations until two agree.

 

[Deleted member 4993]

The problem is that I can't get the correct root with the current derivative. Applying Newton's method always gets me at -15.04....
I need to get the other root at around 11.57...

You'll get that answer if try to find C(r) =0

But need to find C'(r) = 0

Then it works fine without converting to cubic.....