Right circular cylinder: find radius, height, given volume

[kjones]

I am trying to find the radius and height of a container with a capacity of 50 in ^3.
I know that the total voume of the cylinder must be 50 in^3, so volume = 50, or pie times radius ^2 = 50 and height = 50/pie times radius ^2.

I need to substitute the expression for height into the equation for area. I have to find the minimum of area (radius) =

. . .(2 times pie times radius [50/(pie times radius times ^2] )

. . . . . . .+

. . .2 times pie times radius ^2. . .=. . .???

Please help.
K

 

[stapel]

Note: The volume of a right circular cylinder with radius "r" and height "h" is given by:

. . . . .V = (pi)(r²)(h)

...not ("pie")(r²). With only the volume given, the height and radius cannot be fixed; they can only be related.

Eliz.

 

[TchrWill]

I am trying to find the radius and height of a container with a capacity of 50 in ^3. I know that the total voume of the cylinder must be 50 in^3, so volume = 50, or pie times radius ^2 = 50 and height = 50/pie times radius ^2.

I need to substitute the expression for height into the equation for area. I have to find the minimum of area (radius) =

. . .(2 times pie times radius [50/(pie times radius times ^2] )

. . .. . .+

. . .2 times pie times radius ^2. . .=. . .???

Please help.

Having read your problem statement over and over, I cannot help but come to the conclusion that you are seeking the container having the minimum surface area for the 50in^3 volume. On that basis:

V = 50 = PiR^2h making h = 50 - PiR^2
A = PiR^2 + =2PiRh
Substituting for h yields A = PiR^2 + 100PiR - 2Pi^2R^3
The first derivitive yields dA/dR = 2PiR + 100Pi - 6Pi^2R^2 = 0
Simplifying, 6PiR^2 - 2R - 100 = 0
Using the quadratic formula,

. . .R = [2+/-sqrt(4 + =2400Pi)]/12Pi which yields two answers.
. . .R = 2.3569 and h = 2.8650 and
. . .R = 2.3233 and h = 2.9739

Taking the average of the two, R = 2.3396 and h = 2.9076
The surface area becomes A = 59.938...in^2
The volume calculates out to 49.99995...in^3

If this is not what you were looking for, please clarify yourneeds in the problem statement.

 

[galactus]

I got something a little different.

Assuming your container has a top and bottom, the surface area is given by:

\(\displaystyle \L\\2{\pi}r^{2}+2{\pi}rh\).....[1]

The volume is \(\displaystyle \L\\{\pi}r^{2}h=50\)....[2]

Solve for h in [2] and sub into [1].

\(\displaystyle \L\\\frac{50}{{\pi}r^{2}}=h\)

\(\displaystyle \L\\2{\pi}r^{2}+2{\pi}r(\frac{50}{{\pi}r^{2}})=2{\pi}r^{2}+\frac{100}{r}\)

Differentiate:

\(\displaystyle \L\\\frac{d}{dr}[2{\pi}r^{2}+\frac{100}{r}]=4{\pi}r-\frac{100}{r^{2}}\)

Set to 0 and solve for r:

\(\displaystyle \L\\4{\pi}r+\frac{100}{r^{2}}=0\)

Multiply by \(\displaystyle r^{2}\)

\(\displaystyle \L\\4{\pi}r^{3}-100=0\)

\(\displaystyle \L\\r=(\frac{25}{\pi})^{\frac{1}{3}}\approx{1.996}\)

You could round to 2

This gives \(\displaystyle \L\\h=\approx{3.993}\)

You could round to 4.

This gives a surface area of 75.13 square units as the minimum.

And please...."pie"?. Give me a break!

 

[skeeter]

I like this "pie" ...

... better than I like this "pi"

 

[kjones]

CUTE ~ ~ ~

That was mean...I do not know how to show the symbol for pie.......sorry to get your taste buds all in an uproar.

K

 

[galactus]

Everyone is just teasing because you spelled 'pi' as 'pie'.

Did you actually think a popular dessert was a part of the Greek alphabet?.

 

[TchrWill]

I got something a little different...

Yes, galactus, your answer for two end covers is right. While kiones did not specify whether the cylinder had a cover or not, I chose to assume no top just to convey the method. It is well known that the cylinder of minimum surface area (having both a top and bottom) for a specified volume is one where the height is equal to the diameter. I just assumed that if kiones wanted two covers, she at least had the method to derive it if so desired.

I like your sequence of steps. Clearly, mine could have been arranged differently..

 

[galactus]

No problem, TchrWill. I assumed kjones wanted ends because of the way (gender unknown, insert appropriate pronoun here) had it formatted.

Namely,

.(2 times pie times radius [50/(pie times radius times ^2] )

. . . . . . .+

. . .2 times pie times radius ^2. . .=. . .???

Good luck kjones. Don't mind the light-hearted ribbing regarding your 'pie' faux pas.