optimization

[gerauchert]

A cylindrical tin boiler of given volume V has a copper bottom and is open at the top. If sheet copper is 5 times as expensive as sheet tin per unit area, find the most economical dimensions (height and radius) for constructing the boiler.

I can't seem to play around with the equations the right way to get what I need. I figure, if y = total cost and p = price per unit:

y = pCh + 5p*pi*r^2

where C = circumference, r = radius and h = height

This way doesnt seem to be helping me out much though in the calculus aspect... any ideas?

[Method]

Was there any other information included with the problem, like maximum cost, total material/volume, or something like that?

[gerauchert]

Nope, it is as I wrote it. Damn calculus...

[Nava2]

Well, my first impression would be to use implicit differentiation on the function in terms of radius and height, but I think that this wants you to make the connection between radius and height that a square is the most efficient way.

Thus making your equation:
C(r) = 7p*pi*r^2

You can easily differentiate that wrt. to r. The `p' is just a constant as the price doesn't vary.

So:
C`(r) = 14*p*pi*r

The most economically friendly version is to just not build it o.O

Idk, this question needs more information..

[gerauchert]

The answer is given in terms of variables...

r = (V/5pi)^1/3
h = (25V/pi)^1/3

Maybe that will help with the mechanism behind it? Still having trouble with it.

[Nava2]

Just rearrange the volume formula, for variables.

Not too tough then. Just derive assuming that the V-olume is constant.

[Killerdou]

y = pCh + 5p*pi*r^2 you could also just write y = p*2pi*r*h + 5p*pi*r^2
now you know V = pi*r^2*h so h = V/pi*r^2

substitute it and you get y = 2V/r^2 + 5pir^2
y'(r) = 0 gets you V/10pi = r^4 thus you get r and you can use that to calculate your height? Might be wrong though :D

[mixster]

V(olume) is a constant and equal to Pi*r^2*h
So, h = V / (Pi*r^2)

Thus, we can remove h from the cost equation:
C(ost) = ((2*Pi*r*V)/(Pi*r^2)) + (5*Pi*r^2)
C = 2V/r + 5*Pi*r^2
Differentiate C in terms of r:
dC/dr = -2V/r^2 + 10*pi*r

We can assume the optimum cost occurs at a turning point, such that dC/dr = 0
=> 0 = -2V/r^2 + 10*pi*r
=> 2V/r^2 = 10*pi*r
=> 2V = 10*pi*r^3
=> V = 5*pi*r^3
=> r^3 = V/(5*pi)
=> r = cbrt(V/(5*pi)) where cbrt is the cubic root

I'm assuming you could use the same process, only finding r in terms of v & h instead of h in terms of r & h