A rectangle ABOD is inscribed into the region bounded by the x-axis, the y-axis and the graph y= cos2x. what is the greatest perimeter that such a rectangle can have?
Please HELP. I greatly appreciate your help.
let the point of contact on the cosine curve be (x,y)
So the perimeter of the rectangle
= P
= 2x + 2y
= 2x + 2cos 2x
dP/dx = 2 - 4sin 2x
= 0 for a max/min of P
4sin 2x = 2
sin 2x = 1/2
I know sin π/6 = 1/2
so 2x = π/6
x = π/12
then y = cos π/6 = √3/2
then the max of P = π/6 + 2(√3/2)
= π/6 + √3
To find the greatest perimeter of a rectangle inscribed in the region bounded by the x-axis, the y-axis, and the graph y = cos(2x), we need to determine the dimensions of the rectangle first.
A rectangle inscribed in this region would have one side on the x-axis and another side on the y-axis. Let's call the length of the side on the x-axis "x" and the length of the side on the y-axis "y". The perimeter of the rectangle would then be P = 2x + 2y.
Now, let's find the points where the rectangle intersects the graph y = cos(2x). When the rectangle intersects the x-axis, the corresponding y-coordinate is 0, so we know that y = 0 when x = (0, x). Similarly, when the rectangle intersects the y-axis, the corresponding x-coordinate is 0, so we know that x = 0 when y = (0, y).
To find the coordinates of the points where the rectangle intersects the graph y = cos(2x), we substitute y = 0 and x = 0 into the equation.
When y = 0:
0 = cos(2x)
cos(2x) = 0
The cosine function is equal to zero at certain values of 2x. These values can be found by solving the equation cos(2x) = 0:
2x = π/2 + πn or 2x = 3π/2 + πn (where n is an integer)
Simplifying, we have:
x = π/4 + πn/2 or x = 3π/4 + πn/2
So, the possible x-coordinates of the points of intersection of the rectangle and the graph are:
x = π/4 + πn/2 and x = 3π/4 + πn/2
Now, let's find the corresponding y-coordinates:
When x = π/4 + πn/2:
y = cos(2(π/4 + πn/2))
y = cos(π/2 + πn)
y = sin(πn)
y = 0 (since sine function is zero at integer multiples of π)
When x = 3π/4 + πn/2:
y = cos(2(3π/4 + πn/2))
y = cos(3π/2 + πn)
y = sin(π/2 + πn)
y = sin(πn)
y = 0 (since sine function is zero at integer multiples of π)
From the above, we can see that the y-coordinates of the points of intersection of the rectangle and the graph are always 0.
Therefore, the rectangle has two sides on the x-axis, with x-coordinates:
x = π/4 + πn/2 and x = 3π/4 + πn/2
Now, to find the side lengths of the rectangle, we subtract the x-coordinates:
x₂ - x₁ = (3π/4 + πn/2) - (π/4 + πn/2)
= (π/2)
= π/2
So the length of each side of the rectangle is π/2.
Since opposite sides of a rectangle are equal, the length and the width of the rectangle are both π/2.
Now, we can find the perimeter of the rectangle:
P = 2(π/2) + 2(π/2) = 2π + 2π = 4π
Therefore, the greatest perimeter that such a rectangle can have is 4π.
I hope this explanation was helpful! Let me know if you have any further questions.