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
Feb 13, 2017

Answer:

Maximum area is #2.244 " unit"^2# (3dp)

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:

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Where #P(alpha,beta)# is the point in Quadrant 1 where the rectangle intersects the curve #y=2cosx#, and #P'(-alpha,beta)# is the corresponding point in quadrant 2.

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 #A#, as a function of a single variable and to maximize the total area, #A# (It won't matter which variable we do this with as we will get the same result). ie we want a critical point of #A# wrt the variable.

As #P# lies on the curve #y=2cosx#, we have:

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

And the total Area is that of a rectangle of width #2alpha# and height #beta#, so:

# A = 2 alpha beta #
# \ \ \ = 2 alpha (2cos alpha) \ \ \ \ \# (from [1] )
# \ \ \ = 4 alpha cos alpha #

We now have the Area, #A#, as a function of a single variable #alpha#, so differentiating wrt #alpha# (using the product rule) we get:

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

At a critical point we have #(dA)/(d alpha) =0 => #

# 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 #alpha =0.860333589 ...#

With this value of #alpha# we have:

# beta = 1.30436924 ... #
# A = 2.24438535 ... #

We can visually verify that this corresponds to a maximum by looking at the graph of #y=A(alpha)#:

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

So maximum area is #2.244 " unit"^2# (3dp)