# 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?

Feb 13, 2017

Maximum area is $2.244 {\text{ unit}}^{2}$ (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: Where $P \left(\alpha , \beta\right)$ is the point in Quadrant 1 where the rectangle intersects the curve $y = 2 \cos x$, and $P ' \left(- \alpha , \beta\right)$ is the corresponding point in quadrant 2.

Let us set up the following variables:

$\left\{\begin{matrix}\alpha & x \text{-coordinate of point "P \\ beta & y"-coordinate of point "P \\ A & "Total Area of inscribed rectangle}\end{matrix}\right.$

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 $y = 2 \cos x$, we have:

$\beta = 2 \cos \alpha \setminus \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$

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

$A = 2 \alpha \beta$
$\setminus \setminus \setminus = 2 \alpha \left(2 \cos \alpha\right) \setminus \setminus \setminus \setminus \setminus$ (from  )
$\setminus \setminus \setminus = 4 \alpha \cos \alpha$

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

$\frac{\mathrm{dA}}{d \alpha} = \left(4 \alpha\right) \left(- \sin \alpha\right) + \left(4\right) \left(\cos \alpha\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus = 4 \left(\cos \alpha - \alpha \sin \alpha\right)$

At a critical point we have $\frac{\mathrm{dA}}{d \alpha} = 0 \implies$

$4 \left(\cos \alpha - \alpha \sin \alpha\right) = 0$
$\therefore \cos \alpha - \alpha \sin \alpha = 0$

In order to solve this equation we use Newton-Rhapson which gives $\alpha = 0.860333589 \ldots$

With this value of $\alpha$ we have:

$\beta = 1.30436924 \ldots$
$A = 2.24438535 \ldots$

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

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

So maximum area is $2.244 {\text{ unit}}^{2}$ (3dp)