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Optimization Problem

  • Thread starter MitsuShai
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  • #1
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Homework Statement





Homework Equations


Volume of cone= (1/3)*pi*r^2*h
Volume of sphere= (4/3)*pi*r^3
Surface area of sphere 4*pi*r^2

The Attempt at a Solution


primary equation is V(cone)= (1/3)pi*r^2*h---> V(cone)= (1/3)pi*(r-h/2)^2*h
constraint: constraint:V(sphere)= (4/3)*pi*r^3
***from pathagorean theorem, I have to find the radius of the cone because the radius of the sphere is not the same as the radius of the cone.
so the radius of the cone is: (r-h/2),
is my primary equation and constraint switched around because I have to get rid of that h and the only way I can is it to use that and put it in terms of r. But I'm suppose to be looking for the maximum volume of the cone...I don't know how to do this..
 
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Answers and Replies

  • #2
tiny-tim
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Hi MitsuShai! :smile:

(have a pi: π and try using the X2 tag just above the Reply box :wink:)
well I think the primary equation is V= (1/3)pi*r^2*h …
oooh :redface: … wrong area of the base circle! :cry:
 
  • #3
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Hi MitsuShai! :smile:

(have a pi: π and try using the X2 tag just above the Reply box :wink:)


oooh :redface: … wrong area of the base circle! :cry:
It's asking for the maximum volume of the cone, so shouldn't the primary equation be that and then you some how use the sphere...
 
  • #4
tiny-tim
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Yes, but r for the base of the cone isn't the same as r for the radius of the sphere. :wink:
 
  • #5
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Yes, but r for the base of the cone isn't the same as r for the radius of the sphere. :wink:
oh ok, I can see that, so what else can I do
do I assume that the radius is half the height and use pathagorean thereom, so I can get a radius of squareroot(r-h) for the cone?
 
  • #6
tiny-tim
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oh ok, I can see that, so what else can I do
do I assume that the radius is half the height and use pathagorean thereom, so I can get a radius of squareroot(r-h) for the cone?
(and have a square-root: √ :wink:)

Yes, use Pythagoras, but no you don't get √(r - h) :redface:

(besides, isn't h larger than r?)
 
  • #7
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(and have a square-root: √ :wink:)

Yes, use Pythagoras, but no you don't get √(r - h) :redface:

(besides, isn't h larger than r?)
oh sorry I get √(r^2 - h^2),right?
ok so my primary equation is: V= (1/3)pi*(r^2-h^2)*h
constraint: Volume of sphere= (4/3)*pi*r^3
or is it the other way around? because I have to get it in terms of r and so I have to get rid of the h so is it like this:

ok so my primary equation is: Volume of sphere= (4/3)*pi*r^3
constraint: V= (1/3)pi*(r^2-h^2)*h
Solve&sub
h= V/[(1/3)pi*(r^2-h^2)]
V= (1/3)pi*(r-h)*V/[(1/3)pi*(r^2-h^2)]
which goes to V=(r-h)*V/(r^2-h^2)
 
  • #8
tiny-tim
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  • #9
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wrooong! :rolleyes:
CA = u
OA = r
OC = h - r

CA² = OA² - OC²
u² = r² - (h - r)²
u² = 2hr - h²

V(cone) = (1/3)πu²h
V(cone) = (1/3)π(2hr - h²)h
V(cone) = (1/3)π(2h²r - h³)

I can't figure out what to do next though
constraint: V(sphere)= (4/3)*pi*r^3
how can I go from here because I am trying to maximize the volume of the cone, but there is an h as a variable when I'm only suppose to have r. I can't use the constraint because that won't get me anywhere but I can't use the volume of the cone equation to do the rest because that is what's suppose to be maximized....
 
  • #10
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Great this is due in a few hours and I have made no progress :(
 
  • #11
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never mind, i got it.
 

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