Calculus optimization-Drainage channel

**Calculus23** · April 13th 2010, 04:30 PM

A drainage channel of trapezoidal cross section is made by welding three flat strips each 6 cm wide, the middle one being horizontal and the other two inclined at equal angles to it. If the area of the cross section is a maximum, how wide is the channel at the top?

**skeeter** · April 13th 2010, 05:45 PM

Originally Posted by Calculus23

A drainage channel of trapezoidal cross section is made by welding three flat strips each 6 cm wide, the middle one being horizontal and the other two inclined at equal angles to it. If the area of the cross section is a maximum, how wide is the channel at the top?

obviously, you've made a sketch, right?

let $\theta$ = angle of incline for the two sides

area of a trapezoid ...

$A = \frac{h}{2}\left(b_1 + b_2\right)$

$h = 6\sin{\theta}$

$b_1 = 6$

$b_2 = 6 + 2(6\cos{\theta})$

substitute in the values to get area as a function of $\theta$ ... then do that optimization thing.

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