# What is the area of the largest rectangle that can be inscribed under the graph of y=2 cos x for -π /2 ≤x ≤π /2?

Maximum area is

I assume that you man bounded by the x-axis also, otherwise the largest rectangle would be unbounded and therefore infinite.

This is a diagram depicting the problem:

Where

Let us set up the following variables:

# { (alpha, x"-coordinate of point "P), (beta, y"-coordinate of point "P),(A, "Total Area of inscribed rectangle") :} #

Our aim is to find

As

# beta = 2cos alpha \ \ \ \ \ ..... [1]#

And the total Area is that of a rectangle of width

# A = 2 alpha beta #

# \ \ \ = 2 alpha (2cos alpha) \ \ \ \ \# (from [1] )

# \ \ \ = 4 alpha cos alpha #

We now have the Area,

# (dA)/(d alpha) = (4alpha)(-sin alpha) + (4)(cos alpha) #

# \ \ \ \ \ \ = 4(cos alpha - alpha sin alpha )#

At a critical point we have

# 4(cos alpha - alpha sin alpha ) = 0 #

# :. cos alpha - alpha sin alpha = 0 #

In order to solve this equation we use Newton-Rhapson which gives

With this value of

# beta = 1.30436924 ... #

# A = 2.24438535 ... #

We can visually verify that this corresponds to a maximum by looking at the graph of

graph{4xcosx [-4, 4, -5.5, 5.5]}

So maximum area is

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To find the area of the largest rectangle that can be inscribed under the graph of ( y = 2 \cos(x) ) for ( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} ), we need to consider the characteristics of the graph and the properties of rectangles.

The largest rectangle will have its base along the x-axis, and its height will correspond to the maximum value of ( 2\cos(x) ) within the given interval. Since ( \cos(x) ) oscillates between -1 and 1, the maximum value of ( 2\cos(x) ) will be 2.

Thus, the height of the rectangle is 2 units.

To find the width of the rectangle (its base along the x-axis), we look for the intervals where ( \cos(x) ) achieves its maximum value of 1 and its minimum value of -1. This occurs when ( x = \frac{\pi}{2} ) and ( x = -\frac{\pi}{2} ) respectively.

The width of the rectangle is the difference between the x-coordinates of these two points:

[ \text{Width} = \frac{\pi}{2} - (-\frac{\pi}{2}) = \pi ]

Now, we can calculate the area of the rectangle using the formula:

[ \text{Area} = \text{Width} \times \text{Height} ]

[ \text{Area} = \pi \times 2 = 2\pi ]

Therefore, the area of the largest rectangle that can be inscribed under the graph of ( y = 2 \cos(x) ) for ( -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} ) is ( 2\pi ) square units.

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The area of the largest rectangle that can be inscribed under the graph of y = 2 cos(x) for -π/2 ≤ x ≤ π/2 is 4 square units.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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