Extrema

**seuzy13** · Oct 23rd 2009, 12:41 PM

1) The surface area of a cell in a honey comb is:

$S = 6hs + \frac{3s^2}{2}\left(\frac{\sqrt{3} - \cos(\theta)}{\sin(\theta)}\right)$

where h and s are positive constants and theta is the angle at which the upper faces meet the altitude of the cell. Find the angle theta ( $\pi/6 <= \theta <= \pi/2$ ) that minimizes the surface area S.

**CaptainBlack** · Oct 25th 2009, 12:57 AM

Originally Posted by seuzy13

1) The surface area of a cell in a honey comb is:

$S = 6hs + \frac{3s^2}{2}\left(\frac{\sqrt{3} - \cos(\theta)}{\sin(\theta)}\right)$

where h and s are positive constants and theta is the angle at which the upper faces meet the altitude of the cell. Find the angle theta ( $\pi/6 <= \theta <= \pi/2$ ) that minimizes the surface area S.

This has either a calculus like minimum in the interval $[\pi/6,\ \pi/2]$ or the minimum is at an end point of the interval.

So differentiate and set the derivative to zero and solve for a zero in the given interval, if there is no such root examine the value $S$ at each end of the interval.

CB

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the surface area of cell in a honeycomb is given by s=6hs 3s^2/2(√3-cos a/sin a)where h and s are positive integers and angle a is the angle at which the upper surface meet the altitude of the cell.find the angle (30<a>90) that minimises the surface are

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