Help needed from those who are good in math

[ceecee79]

I need some help with 2 math problems from my math class. I used to be good in math, and still am, in some areas, like probability, which I can use in poker, but not so much in other areas, like geometry.

I have until Tuesday to turn these in, so if anyone can help, and explain, it would be greatly appreciated.

1. A steel drum in the shape of a right circular cylinder is required to have a volume of 100 cubic feet.

a) express the amount of material needed to make the cylinder as a function of the radius

b) what radius gives the least amount of material?

c) what is the least amount of material needed for such a radius?

2. A sprinkler that sprays water in a circular motion is to be used to water a square garden. If the area of the garden is 920 sq. ft. find the smallest whole number radius that the sprinkler can be adjusted to so that the entire garden is watered. Also determine how much area outside the garden is watered?

If you can help, please do, and don't just give me the answer, please explain as well.

Thanks

ceecee79 Reviewed by ceecee79 on 09-27-2007. Help needed from those who are good in math I need some help with 2 math problems from my math class. I used to be good in math, and still am, in some areas, like probability, which I can use in poker, but not so much in other areas, like geometry. I have until Tuesday to turn these in, so if anyone can help, and explain, it would be greatly appreciated. 1. A steel drum in the shape of a right circular cylinder is required to have a volume of 100 cubic feet. a) express the amount of material needed to make the cylinder as a Rating: 5

[stinkyattic]

Lol...
Am I reading correctly that you would like us to do your homework for you?
I can give you a hint.
On the first one- Minimizing the surface area of a container for a given volume means getting it as close to spherical as possible.
The second- The key to solving it is that the diagonal of the square garden will be the diameter of the circle. Find the dimensions of the garden. It's simple trig from there out.

[ceecee79]

It's simple trig from there out.

Now that's an oxymoron--"simple trig"--lol.

No, I don't want you to do ALL my homework. I did the rest of it, but I just had issues with these 2 problems.

I just thought I would try here before going to a "homework helpsite", because the cannabis users here are SOOO knowledgable. kiss [insert ass here] lol

Just don't get geometry--never have--never will.

[ReUp]

The first one. (a b and c)

This is the formula for the volume
Area of the circle*height=100
(pi*r^2)*h = 100

Solving for Height to get everything in terms of radius.
100/(pi*r^2) = h

Formula for Surface Area(material):
Circumference*height + 2(area of the circle)
h(2r*pi) + 2(pi*r^2)

Plug in your value of h from above.
(100/(pi*r^2))(2r*pi)) + 2(pi*r^2) = SA

Differentiate that. Solve for 0. Find the minimum. Then plug it in.

I think. Been a while and am doing it in my head.




edit: explanations

[ReUp]

Differentiation turns out to be:
(-200/r^2) + 4*pi*r = dSA

Set it to 0
(-200/r^2) + 4*pi*r = 0

Solve for r
0 = 4*pi*r - 200/r^2
0 = pi*r - 50/r^2
0 = (pi*r^3 - 50)/r^2
0 = pi*r^3 - 50
r^3 = 50/pi
r = (50/pi)^(1/3)

[ReUp]

Now this part I'm not typing out all the steps because its annoying. But you have the radius but you need the height. So you solve this:
h = 100/(pi*r^2)
h = (100/pi)/((50/pi)^(1/3))2

Take my word for it it breaks down into:
h = 2 * (50/∏)^(1/3)
h = 2 * r
h = 2r

Then plug it all back into the original and simplify:
M = 2*pi*r*h + 2*pi*r^2
M = 4*pi*r^2 + 2*pi*r^2
M = 6*pi*r^2
M = 6pi[(50/∏)^(1/3)]2

And thats it.

Enjoy doing your second problem.

[thcbongman]

Dude, I'm 24 and that looks like gibberish to me!

[ceecee79]

Thanks 2 ReUp 4 the info. I am going to make an attempt at understanding it now, I think it may be possible with your information, and the TI-83.

Thanks to stinkyattic 4 the hints.

[ReUp]

The 2nd one is really really easy.
x^2 = 920
x = 31 or something
920 + 920 = c^2
C= square root of 1840 = also the diameter
radius = square root of 1840 divided by 2
pi * r^2 = total area sprinkled
total area sprinkled minus 920 is your answer to the 2nd part.

Sorry I wrote this up real quick its not extremely clear. Ask any questions you have and I'll clarify.

[DemoCommando]

I need some help with 2 math problems from my math class. I used to be good in math, and still am, in some areas, like probability, which I can use in poker, but not so much in other areas, like geometry.

I have until Tuesday to turn these in, so if anyone can help, and explain, it would be greatly appreciated.

1. A steel drum in the shape of a right circular cylinder is required to have a volume of 100 cubic feet.

V=(pi)(radius^2)(height) so substitute for Volume as 100 and
100 = (pi)(radius^2)(height)

a) express the amount of material needed to make the cylinder as a function of the radius

All your doing here is solving for variables: r(adius) or h(eight)
100
(pi)(h) = r(adius) ^2

With the answer you get from above, square root the answer, and that is the function for r.

b) what radius gives the least amount of material?
Use your brains and calculate it out just punching numbers basically to solve for the V=100

c) what is the least amount of material needed for such a radius?
See B

2. A sprinkler that sprays water in a circular motion is to be used to water a square garden. If the area of the garden is 920 sq. ft. find the smallest whole number radius that the sprinkler can be adjusted to so that the entire garden is watered. Also determine how much area outside the garden is watered?
Square garden. So a square has 4 equal sides, and the equation for the area of a square is s(ide) squared.

Area of a circle equation : (pi)(radius squared)

If you can help, please do, and don't just give me the answer, please explain as well.

Thanks

See Above