# Absolute Min & Max

1. Nov 24, 2015

### Michele Nunes

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.

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2. Nov 25, 2015

### FreezingFire

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! :)

3. Nov 25, 2015

### Staff: Mentor

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