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?

forming isosceles trapezoid largest cross section

Answer

optimization cross section isosceles trapezoidLet’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`.