Results 1 to 14 of 14

Math Help - Surface Area Problem.

  1. #1
    Member
    Joined
    Nov 2010
    Posts
    102

    Surface Area Problem.

    In a beehive, each cell is a regular hexagonal prism, open at one end with a trihedral angle at the other end. It is believed that bees form their cells in such a way as to minimize the surface area for a given volume, thus using the least amount of wax in each cell construction. Examination of these cells has shown that the measure of the apex angle theta is amazingly consistent. Based on the geometry of the cell it can be shown that the surface area S is given by

    S=6sh-(3/2)s^2*cot(theta)+[(3s^2*(3)^.5)/2]*csc(theta)

    where s, the length of the sides of the hexagon, and h, the height, are constants.

    a) Calculate ds/d(theta)

    b) What angle should the bees prefer?

    c) Determine the minimum surface area of the cell in terms of s and h.
    Last edited by mr fantastic; December 1st 2010 at 04:03 AM. Reason: Title.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2009
    Posts
    177
    a) Differentiate with regards to \theta.
    b) Find the angle that gives the smallest surface area (use the result from a).
    c) Use the result from b.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by wair View Post
    a) Calculate ds/d(theta)
    do you mean \frac{ds}{d\theta} or \frac{dS}{d\theta} ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2010
    Posts
    102
    I'm sorry I mean dS
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2010
    Posts
    102
    Mondreus do I need to do implicit differentiation ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Nov 2009
    Posts
    177
    No. S is the function and s is simply a constant.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2010
    Posts
    102
    Oh i see thank you
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2010
    Posts
    102
    OK so for part a I get, dS/d(theta) = (3/2)s^2 * csc^2(theta) - [(3s^2)(3)^.5]/2*csc(theta)cot(theta). can someone confirm if this is right ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Nov 2010
    Posts
    102
    For part B would I set my derivative to zero and find the min like any max min problem ? And then for part C it says find the minimum surface are in terms of s and h, but my derivative doesn't have an h.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Nov 2009
    Posts
    177
    I will check a) if you post the original problem and your solution in LaTeX code. Your approach to b) is correct. As for c), it is clearly stated that s and h are constants, so you simply exchange \theta for the angle you found in b) (in the S function).
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Nov 2010
    Posts
    102
    S= 6sh-\frac{3}{2}s^2\cot\theta+(\frac{3s^2\sqrt3}{2})\cs  c\theta

    \frac{dS}{d\theta} = \frac{3}{2}s^2\csc^2\theta-(\frac{3s^2\sqrt3}{2})\csc\theta\cot\theta
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Nov 2009
    Posts
    177
    That looks correct. You can also check your answer here: webMathematica Explorations: Step-by-Step Derivatives
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Member
    Joined
    Nov 2010
    Posts
    102
    Ok that's good. One more question, for part c I'm assuming I plug in the value of theta that I get into the original surface area function because there is no h term in my derivative. correct ?
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Nov 2009
    Posts
    177
    Yes. It's a standard minimization problem.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Surface area problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2011, 05:23 AM
  2. Least Surface Area Problem
    Posted in the Geometry Forum
    Replies: 7
    Last Post: April 13th 2010, 10:44 PM
  3. Surface area problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 14th 2010, 09:38 PM
  4. Surface Area problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 2nd 2010, 03:02 PM
  5. Surface Area Box Problem
    Posted in the Algebra Forum
    Replies: 2
    Last Post: October 19th 2009, 10:00 AM

Search Tags


/mathhelpforum @mathhelpforum