# 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?

##### 1 Answer

#### Answer:

Maximum area is

#### Explanation:

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