Solve Max Volume of Trapezoidal Prism | Optimization Problem

[jen333]

Hi! I have a tricky optimization question here. Not so much setting it up, but solving the derivative for x. (if i did set it up right)

Question: A what angle of theta is the trough (trapezoidal prism) to allow maximum volume?
Unfortunately, i don't have any program that allows me to draw out the diagram (not even paint) so hopefully my symbols/letters diagram will substitute

Face of prism, prism is 20m in length
_______
\ |____| /

the shorter base (b1) is 1m, while the hypotenuse of the end triangles are 1m as well
height is assigned as x while the shorter side of the triangle is (1-x^2)^(1/2)
(the longer base) b2=1+2(1-x^2)^(1/2 theta is the angle between x and the hypotenuse which i know can be solved by cos(theta)=x

so: V=lh (b1+b2)
2
therefore: V=20x(2+2(1-x^2)^(1/2))/2

V'=-20x^2+20(1-x^2)^(1/2)+20-20x^2
0= -40x^2+20(1-x^2)^(1-2)+20

IF i did everything correctly, my question is, how would I solve for x with the root (1-x^2) in the way? I tried squaring all terms, bu that only left me with X^4 and X^2.

help appreciated! thanks.

 

[EnumaElish]

Write V as 0 = 20 (1 - 2 x^2 + Sqrt[1 - x^2])/Sqrt[1 - x^2]

Solve 0 = 1 - 2 x^2 + Sqrt[1 - x^2] by "trial and error." Perhaps try graphing it...

Or, use the foolproof http://en.wikipedia.org/wiki/Quartic_equation#Summary_of_Ferrari.27s_method