# What is the maximum area of a rectangle that can be circumscribed about a given rectangle with length L and width W?

##### 3 Answers

#### Explanation:

Let us set up a concrete scenario...

Start with a rectangle with vertices:

#(L/2, W/2)#

#(-L/2, W/2)#

#(-L/2, -W/2)#

#(L/2, -W/2)#

Rotate it anticlockwise about the origin by

#(L/2 cos theta - W/2 sin theta, W/2 cos theta + L/2 sin theta)#

#(-L/2 cos theta - W/2 sin theta, W/2 cos theta - L/2 sin theta)#

#(-L/2 cos theta + W/2 sin theta, -W/2 cos theta - L/2 sin theta)#

#(L/2 cos theta + W/2 sin theta, -W/2 cos theta + L/2 sin theta)#

(For simplicity, just consider

Then the circumscribing rectangle with sides parallel to the

#(L cos theta + W sin theta)(W cos theta + L sin theta) = (L^2+W^2)cos theta sin theta+WL#

#color(white)((L cos theta + W sin theta)(W cos theta + L sin theta)) = 1/2(L^2+W^2)sin 2theta+WL#

This takes its maximum value when

So the maximum area is:

#WL+1/2(L^2+W^2) = 1/2(L+W)^2#

Unsurprisingly, this is when the circumscribing rectangle is a square.

#### Explanation:

Now using the Lagrange Multipliers technique.

The circumscribed rectangle has the side dimensions

The restrictions are

The lagrangian is

The stationary points are the solutions of

or

Solving for

and the maximum area is

Of course the minimum area circumscribing rectangle has the area

#### Explanation:

And now with rotations.

The circumscribed quadrilateral area is given by

so

Now the maximum is at the solution of

giving

so the solution is for

because at this point