Closed cylindrical can has 500 cm^3 vol, height h, radius r

B

blackroserei

New member
Joined
Jan 20, 2009
Messages
10
Hi I am a current twelfth grader in the IB program (math studies 2) and I'm finding trig and calc extrememly challening. More specifically, its this one problem that is giving me grief.

"A closed cylindrical can has a volume of 500cm^3 (cubed). The height of the can is h cm and the radius of the base is r cm.
a) Find an expression for the total surface area A of the can, in terms of r
b) Given that there is a minimum value of A for r>0, find this value of r"

I've drawn a diagram, but all it is is a cylinder labeled with r and h. I started out by writing down the surface area equation, 2*pie*r*(r + h) but thats as far as I have gotten. I'm completely stuck and have no idea of where to begin! Please help!!
 
blackroserei said:
… I've drawn a diagram …

… I started out by writing down the surface area equation, 2*pie*r*(r + h) …

… have no idea of where to begin! …
Click to expand...


Yes, you do!

You drew a picture, and you wrote an expression for the total surface area.

This is a good start.

(Pi is spelled without an e.)

Part (a) requires you to rewrite the above expression in terms of r only, so you need to express h in terms of r, as well.

In other words, the only variable allowed in the expression is r.

Use the formula for the volume of a cylinder, along with the given volume of 500, to express h in terms of r.

Replace the h with this new expression, and you're done with part (a).

 
Re: HELP!!!

blackroserei said:
Hi I am a current twelfth grader in the IB program (math studies 2) and I'm finding trig and calc extrememly challening. More specifically, its this one problem that is giving me grief.

"A closed cylindrical can has a volume of 500cm^3 (cubed). The height of the can is h cm and the radius of the base is r cm.
a) Find an expression for the total surface area A of the can, in terms of r
b) Given that there is a minimum value of A for r>0, find this value of r"

I've drawn a diagram, but all it is is a cylinder labeled with r and h. I started out by writing down the surface area equation, 2*pie*r*(r + h) but thats as far as I have gotten. I'm completely stuck and have no idea of where to begin! Please help!!
Click to expand...

You did well (except calling ? "a pie" - you should write it as "pi" - or if you are using windows then typing <alt>227 will type ?)

A = 2*?*r*(r + h)

Volume of a cylinder = V = ?*r[sup:1ml7f3ph]2[/sup:1ml7f3ph]*h

500 = ?*r[sup:1ml7f3ph]2[/sup:1ml7f3ph]*h

h = 500/(?*r[sup:1ml7f3ph]2[/sup:1ml7f3ph])

That's the first part - next you need to find 'r' for minimum A

Now what do you want to do...
 
Subhotosh Khan said:
… Now what do you want to do [?]
Click to expand...


Hopefully, use calculus to find the first derivative of the total area expression found in part (a).

Hopefully, set that derivative equal to zero, and solve for r to find the answer to part (b).

Hopefully, show work, if more help is needed.

 
Thanks so much for the help!

Here is what I have done so far...
First I solved for h using the info about the volume and got 500/?r^2 (as someone said to do) and then plugged that in for h in the surface area equation and got 2?r(r+ 500/?r^2)

For part b, first I distributed the 2?r and got the equation 2?r^2 + 1000?r/?r^2. Then I canceled the ?r for the second part of the equation and got 2?r^2 + 1000/ r (I'm not sure if I did that cancellation correctly). I rearranged it so the equation read 2?r^2 + 1000r^-1.

Then I found the derivative and got 4?r - 1000. Then I set the equation equal to zero, solving for r, and got r= 250/?.

I'm not sure if I did all the steps correctly and if the answer is correct.
 
blackroserei said:
Thanks so much for the help!

Here is what I have done so far...
First I solved for h using the info about the volume and got 500/?r^2 (as someone said to do) and then plugged that in for h in the surface area equation and got 2?r(r+ 500/?r^2)

For part b, first I distributed the 2?r and got the equation 2?r^2 + 1000?r/?r^2. Then I canceled the ?r for the second part of the equation and got 2?r^2 + 1000/ r (I'm not sure if I did that cancellation correctly). I rearranged it so the equation read 2?r^2 + 1000r^(-1).

Everything up to this point is correct !

Then I found the derivative and got 4?r - 1000 …

You made a mistake differentiating the second term. Check your work.
Click to expand...
 
Sorry but I'm stuck where I've done something wrong with the second term. Would it be

-1000r^(-2) ?
 
Okay, I understand that part but where to go from there, I'm confused. I now have
0=4pi (r) - 1000r^-2.
But how in the world do I solve that?
 


Please, put parentheses around negative exponents, like I showed you.

Most people learn how to solve equations by taking algebra and pre-calculus before enrolling in a calculus class.

4 * Pi * r - 1000 * r^(-2) = 0

1) Add 1000 * r^(-2) to both sides.

2) Multiply both sides by r^2.

3) Divide both sides by 4 * Pi.

4) Take the cube root of both sides.

5) Simplify the radical.

 
Re:

mmm4444bot said:


Please, put parentheses around negative exponents, like I showed you.

Most people learn how to solve equations by taking algebra and pre-calculus before enrolling in a calculus class.

4 * Pi * r - 1000 * r^(-2) = 0

1) Add 1000 * r^(-2) to both sides.

2) Multiply both sides by r^2.

3) Divide both sides by 4 * Pi.

4) Take the cube root of both sides.

5) Simplify the radical.
Click to expand...

Actually I didn't choose to enroll into a calculus class. Because of requirements of the IB program, Math Studies, which includes aspects of calculus, is required. None of the math classes were chosen. I really do appreciate your help and your willingness to go through these problems step by step for someone you don't even know but the biting sarcasm doesn't help for most people when they're trying to understand something, no matter how slow they are when it comes to math.

Again thank you, and I'll be checking my answers with my math teacher tomorrow.
 
blackroserei said:
I got the answer r = 3?[250/ pi]. Is this right?
Click to expand...


In a class where they touch on "aspects of calculus", it's probably okay.

Since they skipped "aspects of algebra", I'm guessing that they don't care about simplifying radicals, either.

 
Actually the answer was okay and was the exact one my teacher got. My teacher told me it wasn't pertinent to simplify the radical because to take time to unnecessarily simplify something that is already correct can be lethal on a timed exam.

Again, I really appreciate your help but just because I need help in math doesn't mean that I'm stupid regarding underhand insults. Please, your negative comments aren't necessary and distracting from the true purpose of this forum- to help (which you have surely done), not to degrade one who needs help through demeaning comments.
 
blackroserei said:
… it wasn't pertinent to simplify the radical because to take time to unnecessarily simplify …
Click to expand...


After we understand what is necessary and what is not necessary, things flow more smoothly.

As an aside, it takes about 15 seconds to write r = 5 * cuberoot(2/Pi).

MY EDIT: fixed transposition error
 
You must log in or register to reply here.
Top