Results 1 to 2 of 2

Math Help - Optimization

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    23

    Optimization

    Hey there. I basically understand the whole method behind these kinds of problems but on one question in particular I am stuck as to how to set up the equation.

    There is a solid formed by joining a hemisphere (half a sphere) to each end of a right circular cylinder. The Volume must be 3000 feet. The hemispherical ends cost twice as much as the surface area of the sides. i have to find the dimensions that minimize the cost.

    For V i set up the equation v= 4/3(pi)(r^2) + (pi)(r^2)(h)

    Is this right? If it is, what variable do i set the equation equal to, h or r?

    Finally, can someone explain how to come up with the cost equation?
    After that I can take it from there, hopefully. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by painterchica16 View Post
    Hey there. I basically understand the whole method behind these kinds of problems but on one question in particular I am stuck as to how to set up the equation.

    There is a solid formed by joining a hemisphere (half a sphere) to each end of a right circular cylinder. The Volume must be 3000 cubic feet. The hemispherical ends cost twice as much as the surface area of the sides. i have to find the dimensions that minimize the cost.

    For V i set up the equation v= 4/3(pi)(r^3) + (pi)(r^2)(h)

    Is this right? If it is, what variable do i set the equation equal to, h or r?

    Finally, can someone explain how to come up with the cost equation?
    After that I can take it from there, hopefully. Thanks
    Your considerations are correct so far.

    According to the problem you know:

    \dfrac43 \pi r^3 + \pi r^2 h = 3000 ...... [1]

    Now calculate the surface area:

    a=\underbrace{\underbrace{2 \pi r h}_{cylindrical} + \underbrace{4 \pi r^2}_{spherical}}_{parts} ...... [2]


    Costs:

    c = 2\pi r h + 2 \cdot  4 \pi r^2 ...... [3]

    Calculate h from [1] and plug in this term into [3]. You'll get a function of the costs wrt r:

    c(r) = 2\pi r \left( \dfrac{3000}{\pi r^2} - \dfrac43r \right) + 2 \cdot  4 \pi r^2

    Now determine the minimum of c(r).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. optimization help!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 12th 2009, 12:54 AM
  2. Optimization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 8th 2009, 02:09 PM
  3. Optimization
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 29th 2009, 10:56 AM
  4. optimization
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 12th 2008, 10:47 AM
  5. Optimization
    Posted in the Pre-Calculus Forum
    Replies: 0
    Last Post: October 13th 2008, 06:44 PM

Search Tags


/mathhelpforum @mathhelpforum