Calculus - maximum problem
posted by calculus on .
An isosceles trapezoid is inscribed under the graph oof y =cos x. what are the dimensions which will give the greatest area?
I will assume you want the base to be on the x-axis.
Let's make the base as long as possible, that is, from (-pi/2,0) to (pi/2,0)
Let the other point of contact be (x,y)
Area of trap = (1/2)(pi + 2x)(y)
= (1/2)(pi+2x)cosx = (1/2)picosx + xcosx
d(Area)/dx = (-1/2)pi(sinx) + x(-sinx) + cosx
= 0 for a max/min of Area
-pi(sinx) - 2xsinx + 2cosx = 0
2cosx = sinx(pi + 2x)
2/(pi + 2x) = tanx
I ran this through a primitive Newton's Method program and got
x = .45797