anonymous
  • anonymous
A hollow cylinder, open at one end, is constructed of thin sheet metal. The total external surface area of the cylinder is 192pi cm^2. The cylinder has a radius of r cm and a height of h cm. (i) express h in terms of r and show that the volume, Vcm^3 of the cylinder is given by V=pi/2(192r-r^3).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
i give up... sam's latex speed is even faster than my typing!
.Sam.
  • .Sam.
Thats area \[A=\pi r^{2}+2\pi r h=192\pi\]
anonymous
  • anonymous
Yes, I thought so.. I was just about to ask :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

.Sam.
  • .Sam.
Find h from \[A=πr^{2}+2πrh=192π\]
anonymous
  • anonymous
Oh. Ok.
anonymous
  • anonymous
192pi-pi r^2 over 2 pi r =h
.Sam.
  • .Sam.
h should be \[h=\frac{192-r^{2}}{2r}\]
anonymous
  • anonymous
Yes
.Sam.
  • .Sam.
\[V = πr^{2}h\] \[V = πr^{2}\frac{192-r^{2}}{2r}\]
anonymous
  • anonymous
Ok. Thanks. There's a second part of the question. Should I post it here?
.Sam.
  • .Sam.
post here
anonymous
  • anonymous
Given that r can vary, (ii) find the value of r for which V has a stationary value.
anonymous
  • anonymous
I mean V=0?
anonymous
  • anonymous
Is that what I have to do?
.Sam.
  • .Sam.
you need to differentiate V then set dV/dr=0
anonymous
  • anonymous
So, how do I do that?
.Sam.
  • .Sam.
\[V=πr^{2}h\] \[\frac{dV}{dr}=~~~96\pi-\frac{3\pi}{2}r^{2}~~=0~\] \[96\pi-\frac{3\pi}{2}r^{2}~~=0\] \[96\pi=\frac{3\pi}{2}r^{2}~~\] \[\frac{3}{2}r^{2}=96\] \[r^{2}=64\] r=8
anonymous
  • anonymous
Wait, when you differentiate 96pi, doesn't that equal 0?
.Sam.
  • .Sam.
\[V=\frac{\pi}{2}(192r-r^3)\] \[V=96\pi r-\frac{\pi r^{3}}{2}\] \[\frac{dV}{dr}=96\pi -\frac{3\pi r^{2}}{2}\]
anonymous
  • anonymous
Oh, Ok. Silly me!
anonymous
  • anonymous
The last part of the question says find this stationary value and determine whether it is maximum or minimum? Do I differentiate it again?
.Sam.
  • .Sam.
r=8, substitute into \[V=96\pi r- \frac{\pi r^{3}}{2}\] \[V=512\pi\]
.Sam.
  • .Sam.
for max or min, differentiate again
anonymous
  • anonymous
would it be -3pi r = -3 pi times 8 = <0 = maximum point?
.Sam.
  • .Sam.
yes
anonymous
  • anonymous
Thank you :)!

Looking for something else?

Not the answer you are looking for? Search for more explanations.