maximizing volume through change in angle theta

[SilentSymphony]

a trough is made with dimensions 20' long, 1'wide, and 1 ' deep. only angle theta can be varied. what value of theta will maximize the trough's volume?


the picture of the side shows an upside down trapezoid, with a (short end) top length of 1, sides length of 1, and an unknown base length. The angle theta is found on the angles connecting the top of the trapezoid to the sides.

 

[galactus]

I am going to take a guess and say this is your trough:

If so, the area of the trapezoid is \(\displaystyle \L\\A=\frac{1}{2}h(1+b)\)

But, \(\displaystyle \L\\h=sin({\theta})\)

and \(\displaystyle \L\\b=1+2cos({\theta})\)

Therefore, you have: \(\displaystyle \L\\\frac{1}{2}sin({\theta})(2+2cos({\theta}))=sin({\theta})(cos({\theta})+1)\)

Now, differentiate, set to 0 and solve for theta. Where \(\displaystyle 0<{\theta}<\frac{\pi}{2}\)

After you have theta, you can use it in the area formula to find the volume. Don't forget to multiply by 20.