Find the angle 0≤𝜃≤𝜋 0 ≤ θ ≤ π that maximizes the area of the trapezoid with a base of length a = 14 and sides of length b = 2 as in the figure below.

Question

Find the angle 0≤θ≤π that maximizes the area of the trapezoid with a base of length a=14 and sides of length b=2 as in the figure below.

Answer 1

Draw height, h, perpendiculat to base at θ.

Label angle between side and height α.

θ = 90° + α

x represents portion of top base between height and side

x = 2 sin α

Upper base is 14 + 2 sin α

h = 2 cos °

A = (0.5)(14 + 14 + 2 sin α)(2 cos α)

A = (0.5)(28 + 2 sin α)(2 cos α)

A = (14 + sin α)(2 cos α)

Your turn, determine A' by Product Rule, set equal to 0, and solve for α. Remember to add 90° to arrive at θ.

Answer 2

Hi Cesar T.,

What a fine question! So, simply set up the height of the trapezoid as a function of .theta. (it's a trig function), then use that to set up an equation for the area in terms of .theta. (and the known value of a ). Then, since you know that IF there is a maximum, the derivative of that area function with respect to .theta. must be zero (why?) at that "point", find that value (if it exists), evaluate the area there, and compare it with .theta.=.pi./2 case (the rectangle). That will tell you if you have a local minimum or maximum, or if .theta. = .pi./2 is it.

-- Cheers, --Mr. d.