# Thread: Optimization Problem - Trough

1. ## Optimization Problem - Trough

A trough is to be made by bending a long rectangular piece of tin L units wide. The cross section of the trough is an isosceles trapezoid with sides making angles of 120 degree with the base. Find the length of one of the sides that is bent which give a maximum capacity. I don't really know where to begin.

2. Hello, dalbir4444!

Did you make a sketcth?

A trough is to be made by bending a long rectangular piece of tin $L$ units wide.
The cross section of the trough is an isosceles trapezoid with sides
making angles of 120° with the base.
Find the length of one of the sides that is bent which give a maximum capacity.
Code:
      :    ½x     E   L-2x    F    ½x     :
A * - - - - - * - - - - - * - - - - - * B
*         |           |         *
*       |        _  |       *
x *     |      ½√3x |     * x
*30°|           |30°*
* |           | *
D *-----------* C
:   L-2x    :

We have isosceles trapezoid $ABCD$.

Let the two slanted side have length $x\!:\;\;AD = BC = x$

Three of its sides, $AD, DC, CB$ total $L$ units,
. . so: . $DC \:=\:L-2x$

Since $\angle ADC = \angle BCD = 120^o$, then: . $\angle ADE = \angle BCF = 30^o$

$\Delta BFC$ is a right triangle with a 30° angle and hypotenuse $x.$
Then: . $FB\text{ (opp side)} \,=\,\tfrac{x}{2}$ . . . and: . $FC \text{ (adj side)} \,=\,\tfrac{\sqrt{3}}{2}x$

We have a trapezoid:

. . one base is: . $b_1 \:=\:L-2x$

. . the other base is: . $b_2 \:=\:(L-2x) + \tfrac{x}{2} + \tfrac{x}{2} \:=\:L - x$

. . the height is: . $h \:=\:FC \:=\:\tfrac{\sqrt{3}}{2}x$

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

We have: . $A \:=\:\frac{\frac{\sqrt{3}}{2}}{2}\bigg[(L-2x) + (L-x)\bigg] \quad\Rightarrow\quad A \;=\;\tfrac{\sqrt{3}}{4}(2Lx - 3x^2)$

Maximize $A\!:\quad A' \;=\;\tfrac{\sqrt{3}}{4}(2L - 6x) \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{L}{3}}$

The width of the tin sheet is bent into thirds.

3. Thank You! Very helpful solution.