DanJS
  • DanJS
maximize the volume, from a given surface area... @katie00
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
??
DanJS
  • DanJS
that second practice prob you liked me
DanJS
  • DanJS
What is the maximum volume for a sealed cylinder, if the surface area is set at 600pi

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DanJS
  • DanJS
for a cylinder , volume and surface area \[V = \pi*r^2*h\] \[S.A = 2*\pi * r^2 + 2*\pi*r*h\]
DanJS
  • DanJS
given S.A is 600pi, what is the max possible value for the Volume...
DanJS
  • DanJS
One thing to think of, the graph of the function for A ' , the first derivative of Volume will show you when it is increasing/decreasing, depending on the sign of the derivative
DanJS
  • DanJS
Volume and S.A. are both functions of height, and radius, cant do the derivative of A with that
DanJS
  • DanJS
\[S.A = 600\pi = 2*\pi * r^2 + 2*\pi*r*h \] you can solve that surface area either for r or h, solving for h will be easier, since not roots to deal with h = (300/r) - r
DanJS
  • DanJS
put that h in terms of r, intto the Volume Function, substitute \[V = \pi*r^2*[\frac{ 300 }{ r } - r]\]
DanJS
  • DanJS
now you can differentiate that and get the rate of change of the Volume with respect to the radius \[V(r) = 300*\pi*r - \pi*r^3\] and \[V '(r) = 300\pi - 3*\pi*r^2\]
DanJS
  • DanJS
V ' will = 0 at r = 10
DanJS
  • DanJS
before r=10, 0
DanJS
  • DanJS
Overall - You want maximum Volume, that keeps in line with the surface area staying at 600pi 'optimization probs'
DanJS
  • DanJS
1 - put the volume in terms of one variable, we did r 2 - differentiate 3 - the maximums change from increasing functions to decreasing, that occurs at r = 10
DanJS
  • DanJS
most probs are like that process, just different things, boxes, and whjatever
anonymous
  • anonymous
Ty I think I understadn this one much better than the previous one you did. I was working on the previous one and it's still kind of confusing.
DanJS
  • DanJS
oh, can you link me back there
DanJS
  • DanJS
or we can start again new on it
anonymous
  • anonymous
let's start a new one on that one plz!
anonymous
  • anonymous
@DanJS would there be a grpah to this problem?

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