A rectangle is to be inscribed under the arch of the curve y=4cos(.5x) from x=-pi to x=pi. What are the dimensions of the rectanlge with the largest area and what is the largest area? please explain.
The curve describes the arch of a cosine function that goes from 0 at x = pi, to 1 at x = 0, and back down to 0 at x = pi. Pick two points on that curve, at x = + and - a, to construct an inscribed rectangle. The dimensions of the rectangle will be width = 2a and height = 4 cos (a/2).
The area will be A(a)= 8 a cos (a/2).
Find the value of a that maximizes this area
dA/da = 0
8 cos (a/2) -8a sin (a/2) = 0
a = cot (a/2)
That will have to be solved by iteration or graphing. I get a = 1.306
To find the dimensions of the rectangle with the largest area that can be inscribed under the curve, we can use the concept of optimization using calculus.
Step 1: Find the derivative of the area function.
Let's denote the length of the rectangle as L and the width as W. The area function of the rectangle is given by A = LW.
Since we need to maximize the area, we can differentiate the area function with respect to L and W separately to find the critical points.
dA/dL = W
dA/dW = L
Step 2: Set the derivatives equal to zero and solve for L and W.
Since dA/dL = W and dA/dW = L, we have:
W = 0
L = 0
Step 3: Find the critical points.
Given that the length and width must be positive, setting W = 0 or L = 0 will not yield any meaningful solutions. Therefore, there are no critical points.
Step 4: Check the endpoints of the interval.
The curve is defined from x = -π to x = π. We can evaluate the area function at these endpoints.
At x = -π:
A = L * (-2L)
= -2L²
At x = π:
A = L * (2L)
= 2L²
Step 5: Determine the maximum area.
To find the maximum area, we compare the values of the areas at both endpoints and choose the larger one.
Compare -2L² and 2L²:
When L > 0, 2L² > -2L²
When L < 0, 2L² < -2L²
Therefore, the dimensions of the rectangle with the largest area are L = π and W = 2π, and the largest area is A = 2π².