Calculus Application Problem

[clıppy]

Could someone could help me with this question:

A cylindrical aluminium can able to contain half a litre of drink is to be manufactured. The volume of the can must therefore be 500cm³
   a) Find the radius and height of the can which will use the least aluminium and, therefore, be the cheapest to manufacture.
   b) If the radius of the can must be no greater than 5cm, find the radius and height of the can that will use the least aluminium

I've been stumped on the start of the question for an hour now so any help would be greatly appreciated

[kensan]

I'll try and get you started off, this question is a maximization problem. So you will need two equations to work with. Try using the formula for surface area and volume of a cylinder.
See how you go

[clıppy]

Thanks for the help but I'm having trouble understanding how the equation for surface area fits in.

[kensan]

Ok so you have the volume for a cylinder   and also surface area
And you know the volume will be 500. So you let =500
Then you solve for , then sub that value into the equation
Then when you graph that function, you will see the result you are looking for (Use CAS to graph it).
If you need any more assistance, ask

[clıppy]

 
I think I need more assistance

[kensan]

No problem, how far have you got?

[clıppy]

I've solved for h=500/(pi)(r)²
And then subbed that value into the equation for SA which gave me a hyperbola.
At this point I'm not really sure what I'm doing or what I am looking at

[kensan]

Awesome you're on the right track 
So when you look at the graph, you need to realize that you only need to focus on the first quadrant of the graph as we are dealing with a real life situation and the radius cannot be negative. So the x-axis is the value for the radius and the y-axis is the value for the . Using your CAS or algebraic techniques, you can determine where the will be a minimum.
So lets say you do it via CAS, from the graph page, you can find the minimum or maximum of a graph by going Menu/ 6:Analyse Graph/ 2:Minimum. Then you set the boundaries and it gives you where the is a minimum, as well which radius value gives the minimum.

[clıppy]

Ahh okay it's starting to make more sense.
However if we're only focusing on the first quadrant the minimum can continue indefinitely so where are my boundaries?

[kensan]

The equation I ended up with was 
Which means where the derivative=0 is a minimum, maybe you made an error when substituting in?

[clıppy]

We have the same equation just in different forms.
So I'm still confused as to, if it's a hyperbola and we are looking at the first quadrant (top right) as we go across the positive x-axis (radius), the y-axis (SA) keeps decreasing.
So how can a boundary be found?

[kensan]

Ohhh it could be your window settings, I changed my window screen to -1<X<10 and -100<Y<2000
Yeah it's a good skill to be able to adjust the window settings effectively. You can get some graphs that are way off the standard scale.

[clıppy]

Ohhhhhhhh! I see it now, it's not a hyperbola it's a parabola(ish)
Using that I can now see the radius is 4.3 with a surface area of 348 and using that I can find the height. Right?
So if I was supposed to do this all algebraically, how would I go about it?

[kensan]

Yeah you just sub it into the isolated value you would have found earlier.

Ok this is how I would do it by hand.


Differentiate


Let it =0




  (When the numerator = 0, the whole thing =0)\





By CAS you just define the equation, derive, then solve for 0
Fairly sure that we won't be expected to do it all by hand though.

[clıppy]

Perfect.
Thanks a ton kenoy