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Homework Help: Absolute Min & Max

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data
    The surface area of a cell in a honeycomb is S = 6hs + (3s2/2)((√3 - cosθ)/sinθ) where h and s are positive constants and θ is the angle at which the upper faces meet the altitude of the cell. Find the angle θ (π/6 ≤ θ ≤ π/2) that minimizes the surface area S.

    2. Relevant equations

    3. The attempt at a solution
    Okay so first I took the derivative of the equation and the (3s2/2) isn't relevant since there's no θ involved so I'm going to disregard that part. Once simplified, on the numerator I obtained (1 - √3cosθ) and on the denominator just sin2θ but all the thetas that would make the denominator = 0 are not in the given domain of theta so then I focused on the numerator. The theta that would make the numerator = 0 is cos-1(1/√3) which should end up being the theta that gives the minimum value for the surface area, however the back of the book says theta should = (√3)/3 and I'm not sure where they got that value from.

    Attached Files:

  2. jcsd
  3. Nov 25, 2015 #2
    If you mean that the book says ##\large \theta = \cos^{-1}\left(\frac {\sqrt 3}{3}\right)##, then it is the same as ##\large \theta = \cos^{-1}\left(\frac {1}{\sqrt 3}\right)##! Your answer is correct! :)
  4. Nov 25, 2015 #3


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    Staff: Mentor

    \frac{1}{\sqrt{x}} = \frac{1}{\sqrt{x}} \times \frac{\sqrt{x}}{\sqrt{x}} = \frac{\sqrt{x}}{x}
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