The figure below shaded i gray is that of a trapezoid having base length a and slant height b.?

Question

Find the max possible area, Amax, of the trapezoid when a= b= 8 inches.
http://i181.photobucket.com/albums/x85/ctti_3/untitled-30.jpg
i try to work this out but my answer turn out to be 32(root 3) but it wasnt in the answer choices. Can you check and see if it's right?

Answer 1

a=b=8
I will refer to theta as "t"

Height = bcos(t)
Base = a+2b*sin(t)

Area = A = (b/2)cos(t)*(2a+2b*sin(t))
= b*cos(t)(a+b*sin(t))
= 64*cos(t)(1 + sin(t))
dA/dt = 64*[cos^2(t) - sin(t)(1+sin(t))]
dA/dt = 0
cos^2(t) = sin(t) + sin^2(t)
sin(t) = u
1 - u^2 = u + u^2
2u^2 + u - 1 = 0
2u^2 + 2u - u - 1 = 0
2u(u+1) - 1(u+1) = 0
(u+1)(2u-1) = 0
u = -1
2u = 1
u = 1/2

If u = -1 (theta = 270 degrees - not possible)

If u = 1/2 (theta = 30 degrees)
sin(t) = 1/2
cos(t) = sqrt(3)/2

Area = (64*sqrt(3)/2)(1 + 1/2)
= 32sqrt(3) * 3/2
= 48sqrt(3)

Answer 2

Area of trapezium = area of rect plus area of two triangles
= a * b cos theta + b cos theta * b sin theta
if a=b=8 therefore we have
64 cos theta + 64 sin theta cos theta
= 64 cos theta (1+ sin theta).

Max at theta = 30 degrees.
= 48 sqrt(3).

Answer 3

The maximum area occurs when θ = π/6, producing an area of 48*sqrt(3).