## Question

We need to construct a water channel with aluminium material. We have a piece of rectangular aluminium plate with width 24 cm. As shown in Figure, the channel will be formed with a shape of isosceles trapezoid (with an open topside).

How to bend the plate to achieve the largest cross section of the channel?

## Answer

Let’s redraw trapezoid upside down and draw altitudes.

Angle `ACF` is also `Phi`.

Since trapezoid is isosceles then angle `BDG` is also `Phi`.

Let `AC=x` then `BD` also equals `x`.

Since width of plate is 24 then `AB=24-AC-BD=24-x-x=24-2x`.

Next, since `ABGF` is rectangle then `FG=AB=24-2x`.

Now, from right-angled triangle `AFC` `h=AF=BG=xsin(Phi)`.

Also, `CF=xcos(Phi)`. Similarly, `GD=xcos(Phi)`.

So, `CD=CF+FG+GD=xcos(Phi)+24-2x+xcos(Phi)=24-2x+2xcos(Phi)`.

Thus, area of trapezoid is `A=(AB+CD)/(2)*h=(24-2x+24-2x+2xcos(Phi))/(2)*xsin(Phi)`.

`A=(24-2x+xcos(Phi))xsin(Phi)=24xsin(Phi)-2x^2sin(Phi)+x^2cos(Phi)sin(Phi)`.

Simplifying further gives `A=24xsin(Phi)-2x^2sin(Phi)+1/2x^2sin(2Phi)`.

We have following constraints: `x>0` (length of the side should be positive), `0<Phi<pi`.

To find maximum, we find extrema points with the help of derivatives.

For this we need to solve following system:

`{((partial A)/(partial x)=0),(quad),((partial A)/(partial Phi)=0):}`.

`(partial A)/(partial x)=24sin(Phi)-4xsin(Phi)+xsin(2Phi)`.

`(partial A)/(partial Phi)=24xcos(Phi)-2x^2cos(Phi)+x^2cos(2Phi)`.

Now, system can be rewritten as follows:

`{(24sin(Phi)-4xsin(Phi)+xsin(2Phi)=0),(24xcos(Phi)-2x^2cos(Phi)+x^2cos(2Phi)=0):}`.

Since `x!=0` (length can't be zero), we divide second equation of the system by `x`:

`{(4xsin(Phi)-xsin(2Phi)=24sin(Phi)),(24cos(Phi)-2xcos(Phi)+xcos(2Phi)=0):}`.

From first equation `x=(24sin(Phi))/(4sin(Phi)-sin(2Phi))`.

From second equation `x=(24cos(Phi))/(2cos(Phi)-cos(2Phi))`.

Equating right hand sides gives the following:

`(24sin(Phi))/(4sin(Phi)-sin(2Phi))=(24cos(Phi))/(2cos(Phi)-cos(2Phi))`.

Divide left and right part by 24:

`(sin(Phi))/(4sin(Phi)-sin(2Phi))=(cos(Phi))/(2cos(Phi)-cos(2Phi))`.

`sin(Phi)(2cos(Phi)-cos(2Phi))=cos(Phi)(4sin(Phi)-sin(2Phi))`.

`2sin(Phi)cos(Phi)-sin(Phi)cos(2Phi)=4sin(Phi)cos(Phi)-cos(Phi)sin(2Phi)`.

`2sin(Phi)cos(Phi)-cos(Phi)sin(2Phi)+sin(Phi)cos(2Phi)`.

Now, we use formulas for converting product of trigonometric functions into sum:

`2sin(Phi)cos(Phi)-(sin(2Phi-Phi)+sin(2Phi+Phi))/2+(sin(Phi-2Phi)+sin(Phi+2Phi))/2=0`.

`2sin(Phi)cos(Phi)-(sin(Phi)+sin(3Phi))/2+(sin(-Phi)+sin(3Phi))/2=0`.

`2sin(Phi)cos(Phi)-(sin(Phi)+sin(3Phi))/2+(-sin(Phi)+sin(3Phi))/2=0`.

`2sin(Phi)cos(Phi)-sin(Phi)=0`.

`sin(Phi)(2cos(Phi)-1)=0`.

So, either `sin(Phi)=0` or `cos(Phi)=1/2`.

There are no solutions to the equation `sin(Phi)=0` on interval `(0,pi)`.

Equation `cos(Phi)=1/2` has only one solution on interval `(0,pi)`, namely, `Phi=pi/3`.

Now, `x=(24sin(Phi))/(4sin(Phi)-sin(2Phi))=(24sin(pi/3))/(4sin(pi/3)-sin(2pi/3))=(24sqrt(3)/2)/(4sqrt(3)/2-sqrt(3)/2)=24/(4-1)=8`.

To check whether it is maximum we use second partial derivatives test:

`(partial^2 A)/(partial x^2)=-4sin(Phi)+sin(2Phi)`.

`(partial^2 A)/(partial Phi^2)=-24xsin(Phi)+2x^2sin(Phi)-2x^2sin(2Phi)`.

`(partial^2 A)/(partial x partial Phi)=24cos(Phi)-4xcos(Phi)+2xcos(2Phi)`.

`(partial^2 A)/(partial x^2)(8,pi/3)=-4sin(pi/3)+sin(2pi/3)=-3sqrt(3)/2<0`.

`(partial^2 A)/(partial Phi^2)(8,pi/3)=-24*8*sin(pi/3)+2*8^2sin(pi/3)-2*8^2sin(2pi/3)=-96sqrt(3)`.

`(partial^2 A)/(partial x partial Phi)(8,pi/3)=24cos(pi/3)-4*8*cos(pi/3)+2*8*cos(2pi/3)=-12`.

`D=(partial^2 A)/(partial x^2)(8,pi/3)(partial^2 A)/(partial Phi^2)(8,pi/3)-((partial^2 A)/(partial x partial Phi)(8,pi/3))^2=-3sqrt(3)/2*(-96sqrt(3))-(-12)^2=288>0`.

Since `D>0` and `(partial^2 A)/(partial x^2)(8,pi/3)<0` then point `(8,pi/3)` is indeed maximum.

**Answer:** `x=8` cm and `Phi=pi/3=60^0`.