# Calculus optimization-Drainage channel

• Apr 13th 2010, 04:30 PM
Calculus23
Calculus optimization-Drainage channel
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?
• Apr 13th 2010, 05:45 PM
skeeter
Quote:

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

area of a trapezoid ...

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

$\displaystyle h = 6\sin{\theta}$

$\displaystyle b_1 = 6$

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

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