maximum volume/ dimension calc question?

Question

find the dimensions of the trapezoid of maximum area that can be inscribed in a circle of radius R with its base a diameter of the circle.

Answer 1

Let AB be the diameter of the circle and the base of the trapezoid. Let CD be the other base (bl). We will assume ABCD, the trapezoid is isoceles. Let O be the center of the circle, and AOB is a line segment. Draw radii from O to C and D. Now, drop a perpendicular from C to AOB which meets AOB at E, and one from D to AOB which meets AOB at F. Now CD= EO+OF= 2 EO or 2 OF. Call EO=b2. Then b2^2+h^2=R^2.
Now bl = 2b2, and the lower base = 2R The area of the trapezoid is (h/2)(bl + 2R). We express bl in terms of h and R. Since R is a constant, we have the area= f(h)
So you take derivitive d(area)/dh, and set =0, and solve for h.

Answer 2

Hint : the top of the trapezoid touches the circle and all the points on the circle are (x,y) where x^2 + y^2 = R^2. And YOU certainly know the formula for the area of a trapezoid...if not take another copy of it ,invert it and place next to the original...now you have a parallelogram...compute that area and divide by 2.
If R =1 The (x,y) is close to (0.6,0.8) but not the true values...your answers need to be in terms of R .

Answer 3

surface area, s, of box 2x by x by x is s = 2( 2x*x + x*x + x*x) = 216 8x^2 = 216 x^2 = 27 x = 3 sqrt(3) V= 2x^3 : This the maximum and minimum volume for a box 2x by x by x with surface area of 216