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.