# Maximum

• Nov 10th 2008, 12:00 AM
ihmth
Maximum
Need help, I don't know what this trapezoid gutter really look like :(

1.) A trapezoid gutter is to be made from a strip of metal 22 in. wide, by bending up the edges. If the bases is 14 in. wide, what width across the top gives the greatest carrying capacity? (use trig function)
• Nov 10th 2008, 03:34 AM
Soroban
Hello, ihmth!

Quote:

1) A trapezoid gutter is to be made from a strip of metal 22 in. wide, by bending up the edges.
If the base is 14 in. wide, what width across the top gives the greatest carrying capacity?
(use trig function)

Code:

F              E
A * - - * - - - - - - - * - - * D
\    :              :    /
\  :              :  /
4 \  :              h:  / 4
\ :              :θ/
\:              :/
*---------------*
B      14      C

The isosceles trapezoid is $ABCD$, where $BC = 14$, and $AB = CD = 4.$
Let $\theta = \angle ECD$

Area Formula: . $A \;=\;\frac{h}{2}(b_1 + b_2)$

In right triangle $DEC\!:\;\;DE = 4\sin\theta,\;EC = 4\cos\theta$

So we have: . $h \:=\:4\cos\theta,\;b_1 \:=\: 14,\;b_2 \:=\:14 + 8\sin\theta$

Then: . $A \;=\;\frac{4\cos\theta}{2}\bigg[14 + (14 + 8\sin\theta)\bigg] \quad\Rightarrow\quad A \;=\;8\cos\theta(7 + 2\sin\theta)$

. . and that is the function to be maximized . . .