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.

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

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

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

To find the dimensions that give the maximum area, we need to consider the properties of the rectangle inscribed under the curve y = 4cos(0.5x) and its area.

Let's start by visualizing the problem. The curve y = 4cos(0.5x) is a periodic function that oscillates between -4 and 4. Our task is to find the rectangle whose base lies on the x-axis (from x = 0 to x = pi) and has its top corners touching the curve.

To proceed, we'll consider the height and width of the rectangle:

1. Height: The height of the rectangle is equal to the y-coordinate of the points on the curve at x = 0 and x = pi. Let's calculate these values.
At x = 0: y = 4cos(0.5(0)) = 4cos(0) = 4.
At x = pi: y = 4cos(0.5(pi)) = 4cos(0.5pi) = 4cos(90) = 4(0) = 0.

So, the height of the rectangle is 4 units.

2. Width: The width of the rectangle is equal to the difference between the x-coordinates of the two points where the curve intersects the x-axis. These points can be found by setting y = 0 and solving for x.
0 = 4cos(0.5x)
cos(0.5x) = 0

Since the cosine function is zero at multiples of pi/2, we have:
0.5x = pi/2 or 0.5x = 3pi/2

Solving these equations for x, we get:
x = pi/2 or x = 3pi/2

The width of the rectangle is the difference between these x-coordinates:
Width = 3pi/2 - pi/2
= pi

Now that we have the height and width of the rectangle (4 units and pi units respectively), we can calculate the area.

Area = Height x Width
= 4 x pi
≈ 12.57 square units

Therefore, the dimensions that give the maximum area are a height of 4 units and a width of pi units. The maximum area is approximately 12.57 square units.