Originally Posted by

**Archie Meade** Hi Karlos,

The volume of the trough can be subdivided into the central cube

and the two outside pyramids.

To calculate volume, use

$\displaystyle V=(cross-sectonal\ area)(length)$

hence you must express the trough's height "h" in terms of $\displaystyle \theta$

$\displaystyle cos\theta=\frac{h}{1}=h$

The cross-sectional area of the right and left triangles are

$\displaystyle \frac{1}{2}(h)(1)sin\theta=\frac{1}{2}cos\theta\ sin\theta$

Hence the volume of the trough is

$\displaystyle V=20cos\theta+20(2)\frac{1}{2}cos\theta\ \sin\theta=20cos\theta(1+sin\theta)$

Theta giving max volume is found by differentiating this and setting it to zero, using the product rule of differentiation

$\displaystyle \frac{d}{d\theta}\left[20cos\theta(1+sin\theta)\right]$

$\displaystyle 20[cos\theta\ cos\theta+(1+sin\theta)(-sin\theta)]=0$

$\displaystyle 20\left(cos^2\theta-sin\theta-sin^2\theta\right)=0$

$\displaystyle 1-sin^2\theta-sin\theta-sin^2\theta=0$

$\displaystyle 2sin^2\theta+sin\theta-1=0$

$\displaystyle (2sin\theta-1)(sin\theta+1)=0$

For an acute angle use the first factor

$\displaystyle 2sin\theta=1$

$\displaystyle sin\theta=\frac{1}{2}$

$\displaystyle \theta=sin^{-1}\frac{1}{2}=30^o$