A rectangle is to be inscribed under the curve y=4cos(.5x). The rectangle is to be inscribed from x=0 to x=pi. Find the dimensions that give max area and what is the max area.

Answer 1

Length of the rectangle = Distance from origin along x = x co=ordinate = x

Breadth of rectangle = Distance from origin along y = y co-ordinate = 4cos(.5x)

Area = 4cos(.5x) * x = f(x)

This area function is the function to be maximized.

f'(x) = d(x4cos(.5x))/dx
= 4cos(.5x) - 2xsin(.5x)

Equating f'(x) = 0,
4cos(.5x) - 2xsin(.5x) = 0
=> 4cos(.5x) = 2xsin(.5x)
=> 0.5x = cot(0.5x)

I'm not sure how you'd get an exact value for A = cot(A), I'd like to see what the other tutors have in mind

Answer 2

Let the vertices be(x,0), (x,y), (0,y) and (0,0)
then the area is xy
A = x(4cos(x/2))
dA/dx = x(-4sin(x/2)(1/2) + 4cos(x/2)
= 0 for a max of A

x(2sin(x/2) = 4cos(x/2)
sin(x/2)/cos(x/2) = 2/x
tan(x/2) = 2/x

no nice way to solve this, Wolfram says
https://www.wolframalpha.com/input/?i=tan(x%2F2)+%3D+2%2Fx
x = appr 1.72 within our given domain
y = appr 2.61

max area = appr 4.49 units^2

check my algebra

Answer 3

Trying the old fashioned "a bit higher and a bit lower" trick

let x= 1.80 , y = 4cos(1.80/2) = 2.486
area = 4.476 < 4.49

let x=1.6, y = 2.787
area = 4.459 < 4.49

my above answer is reasonable