Finding the lateral area of a cylinder inscribed in a circle

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roxstar1

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For this problem, I need to find the lateral area of a cylinder inscribed in a circle. The diameter of the cylinder is 8 ft.

I am very confused on this problem and none of the techniques I have used have worked. Any help would be greatly appreciated.
 
There must be more information. The diameter of the circle can be anything from 8' to 8000 miles.
Gene
 
must be more information

sorry...I meant to say that the diameter of the circle (not the cylinder) is 8 feet..I hope this helps
 
Sorry, still something missing. Now the diameter of the cylinder can be anything from 0 to 8' with zero area at both ends. Is it possible they want the MAXIMUM lateral area? Why don't you copy the problem exactly as presented!
Gene
 
complete problem

Find the maximum lateral area of a cylinder with no top or bottom, which is inscribed in a circle, given a diameter of 8ft.
 
Now we have it!
Draw a circle with a rectangle inscribed.
Label the top of the rectangle w
Label the side of the rectangle h
w²+h²=8²
Area = \(\displaystyle \pi\)*w*h
Solve the first for
h=sqrt(8²-w²)
substitute for h in the second for
A = \(\displaystyle \pi\)w*sqrt(8²-w²)=
\(\displaystyle \pi\)sqrt(8<sup>2</sup>w<sup>2</sup>-w<sup>4</sup>)

That is what you have to find dA/dw for.

You don't say where you are stuck. Can you take it from there?
 
Could you possibly mean "Find the dimensions of the cylinder of greatest surface area that can be inscribed within a sphere of radius R?."
 
can't find critical number

so the lateral area of the cylinder = 2(pi)rh
then, 8^2 = h^2 + w^2
h = (8^2-w^2)^1/2
h = 8-w
A = 2(pi)(4)(8-w)
A= 64(pi)-8(pi)w
dA/dw = ?

here is where I keep getting stuck...I cant find the critical numbers because "w" goes away when I take the derivitive of A.





Gene said:
Now we have it!
Draw a circle with a rectangle inscribed.
Label the top of the rectangle w
Label the side of the rectangle h
w²+h²=8²
Area = \(\displaystyle \pi\)*w*h
Solve the first for
h=sqrt(8²-w²)
substitute for h in the second for
A = \(\displaystyle \pi\)w*sqrt(8²-w²)=
\(\displaystyle \pi\)sqrt(8<sup>2</sup>w<sup>2</sup>-w<sup>4</sup>)

That is what you have to find dA/dw for.

You don't say where you are stuck. Can you take it from there?
Click to expand...
 
problem

Galatus has a point!
Isn't a cylinder a 3 dimensional object, where as a circle is a 2 dimensional object??
So, its like the old saying "adding apples and oranges"!!!!!!

It must have been the maximum lateral surface of a cylinder inscribed in a sphere!
 
Your problem is this (invalid) step.
h = (8^2-w^2)^1/2
h = 8-w
Suppose w = 4.
The first says h = 6.9...
The second says h = 4
They are not the same. You have to use the first in the area equation
substitute for h...
A = w*sqrt(8²-w²)=
sqrt(8²w^2-w^4)
Click to expand...
When you take dA/dw the w will hang around so you can set the numerator = 0 and solve for w.

PS. In re circle vs sphere. You guys are probably right but...
It doesn't change the problem. The cylinder would stick out of the circles' plane into the third dimension and all the answers would be the same.
 
Galacus: Please explain where your answer is any different than mine except that my w = your 2r. I appreciate corrections or improvements but xeroxes annoy me.
-----------------
Gene
 
not sure im doing this right...

I must be doing something wrong because I am not finding the height to equal the diameter. here's what I have so far:

S = 4(pi)r((16-r^2)^(1/2))

dS/dr = (-4(pi)r^2 + 4(pi)(16-r^2)/((16-r^2)^1/2) = 0

maybe I messed up taking the derivitive...but I tried a few times and kept getting the same thing.
I found the critical number: r = 2(2)^1/2

this r value is a maximum but doesnt make h= diameter

r must be = 0 to make the h= diameter (8ft) right? but then...how can r = 0?
r^2 + (h/2)^2 = 16

I have gotten lost once again...any suggestions?
 
I didn't mean for G to delete his. They are the same. I'll use mine cause its still there.
A=pi*w*sqrt(8²-w²)=
pi*(64w^2-w^4)^(1/2)
dA/dw = pi*(1/2)(128w-4w^3)/(64w^2-w^4)^(1/2)
It doesn't matter what the denominator is (as long as its not zero) so we want
(128w-4w^3) = 0
factoring the w (we aren't interested in w=0) we want
128-4w^2 = 0
w=4sqrt(2)
Substitute w in the area formula.
 
I get it now

oh they are the same...ok...I think I get it! thanks so much for all your help :)
 
The only difference is that where I used w he used w/2 = r. If you wrote his down you should get A = about 100 with either one.
---------------------
Gene
 
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