Hello, dalbir4444!

Did you make a sketcth?

A trough is to be made by bending a long rectangular piece of tin $\displaystyle L$ 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 $\displaystyle ABCD$.

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

Three of its sides, $\displaystyle AD, DC, CB$ total $\displaystyle L$ units,

. . so: .$\displaystyle DC \:=\:L-2x$

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

$\displaystyle \Delta BFC$ is a right triangle with a 30° angle and hypotenuse $\displaystyle x.$

Then: . $\displaystyle FB\text{ (opp side)} \,=\,\tfrac{x}{2}$ . . . and: .$\displaystyle FC \text{ (adj side)} \,=\,\tfrac{\sqrt{3}}{2}x$

We have a trapezoid:

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

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

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

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

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

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

The width of the tin sheet is bent into thirds.