Find Min Surface Area: Absolute Min & Max Homework

[Michele Nunes]

Homework Statement

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.

Homework Equations

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.

 

[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! :)

 

[DrClaude]

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