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

Feb 5, 2017

$\frac{1}{2} {\left(L + W\right)}^{2}$

#### Explanation:

Let us set up a concrete scenario...

$\left(\frac{L}{2} , \frac{W}{2}\right)$

$\left(- \frac{L}{2} , \frac{W}{2}\right)$

$\left(- \frac{L}{2} , - \frac{W}{2}\right)$

$\left(\frac{L}{2} , - \frac{W}{2}\right)$

Rotate it anticlockwise about the origin by $\theta$ to give a rectangle with vertices:

$\left(\frac{L}{2} \cos \theta - \frac{W}{2} \sin \theta , \frac{W}{2} \cos \theta + \frac{L}{2} \sin \theta\right)$

$\left(- \frac{L}{2} \cos \theta - \frac{W}{2} \sin \theta , \frac{W}{2} \cos \theta - \frac{L}{2} \sin \theta\right)$

$\left(- \frac{L}{2} \cos \theta + \frac{W}{2} \sin \theta , - \frac{W}{2} \cos \theta - \frac{L}{2} \sin \theta\right)$

$\left(\frac{L}{2} \cos \theta + \frac{W}{2} \sin \theta , - \frac{W}{2} \cos \theta + \frac{L}{2} \sin \theta\right)$

(For simplicity, just consider $0 \le \theta \le \frac{\pi}{2}$ so we don't have to be concerned about a variety of cases, etc.)

Then the circumscribing rectangle with sides parallel to the $x$ and $y$ axes has area:

$\left(L \cos \theta + W \sin \theta\right) \left(W \cos \theta + L \sin \theta\right) = \left({L}^{2} + {W}^{2}\right) \cos \theta \sin \theta + W L$

$\textcolor{w h i t e}{\left(L \cos \theta + W \sin \theta\right) \left(W \cos \theta + L \sin \theta\right)} = \frac{1}{2} \left({L}^{2} + {W}^{2}\right) \sin 2 \theta + W L$

This takes its maximum value when $\sin 2 \theta = 1$, e.g. when $\theta = \frac{\pi}{4}$

So the maximum area is:

$W L + \frac{1}{2} \left({L}^{2} + {W}^{2}\right) = \frac{1}{2} {\left(L + W\right)}^{2}$

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

Feb 6, 2017

$\frac{1}{2} {\left(L + W\right)}^{2}$

#### Explanation:

Now using the Lagrange Multipliers technique.

The circumscribed rectangle has the side dimensions

$\left(a + b\right)$ and $\left(c + d\right)$ so the sough area is $\left(a + b\right) \left(c + d\right)$

The restrictions are

${a}^{2} + {b}^{2} = {L}^{2}$ and
${c}^{2} + {d}^{2} = {W}^{2}$

The lagrangian is

$\Phi \left(a , b , c , d , {\lambda}_{1} , {\lambda}_{2}\right) = \left(a + b\right) \left(c + d\right) + {\lambda}_{1} \left({a}^{2} + {b}^{2} - {L}^{2}\right) + {\lambda}_{2} \left({c}^{2} + {d}^{2} - {W}^{2}\right)$

The stationary points are the solutions of

$\nabla \Phi = \left({\Phi}_{a} , {\Phi}_{b} , {\Phi}_{c} , {\Phi}_{d} , {\Phi}_{{\lambda}_{1}} , {\Phi}_{{\lambda}_{2}}\right) = 0$

or

$\left\{\begin{matrix}c + d + 2 a {\lambda}_{1} = 0 \\ c + d + 2 b {\lambda}_{2} = 0 \\ a + b + 2 c {\lambda}_{2} = 0 \\ a + b + 2 d {\lambda}_{1} = 0 \\ {a}^{2} + {d}^{2} - {L}^{2} = 0 \\ {b}^{2} + {c}^{2} - {W}^{2} = 0\end{matrix}\right.$

Solving for $a , b , c , d , {\lambda}_{1} , {\lambda}_{2}$ we get at

$\left(\begin{matrix}a = \frac{L}{\sqrt{2}} \\ b = \frac{W}{\sqrt{2}} \\ c = \frac{W}{\sqrt{2}} \\ d = \frac{L}{\sqrt{2}} \\ {\lambda}_{1} = - \frac{L + W}{2 L} \\ {\lambda}_{1} = - \frac{L + W}{2 W}\end{matrix}\right)$

and the maximum area is

$\left(a + b\right) \left(c + d\right) = \frac{1}{2} {\left(L + W\right)}^{2}$

Of course the minimum area circumscribing rectangle has the area $L \cdot W$

Feb 6, 2017

$\frac{1}{2} {\left(L + W\right)}^{2}$

#### Explanation:

And now with rotations.

The circumscribed quadrilateral area is given by

$A \left(\theta\right) = 2 \left(\frac{1}{2} \left(W \sin \theta\right) \left(W \cos \theta\right) + \frac{1}{2} \left(L \sin \theta\right) \left(L \cos \theta\right)\right) + L W$

so

$A \left(\theta\right) = \left({W}^{2} + {L}^{2}\right) \sin \theta \cos \theta + L W$

Now the maximum is at the solution of

$\frac{\mathrm{dA}}{d \theta} = \left({W}^{2} + {L}^{2}\right) \left(1 - 2 {\sin}^{2} \theta\right) = 0$

giving $\theta = \left\{- 3 \frac{\pi}{4} , - \frac{\pi}{4} , \frac{\pi}{4} , 3 \frac{\pi}{4}\right\}$. Between those solutions we will choose the maximum. The local maxima are located at points in which $\frac{{d}^{2} A}{d {\theta}^{2}} < 0$

so the solution is for $\theta = \frac{\pi}{4}$ or $\theta = - 3 \frac{\pi}{4}$

$A \left(\frac{\pi}{4}\right) = \left({W}^{2} + {L}^{2}\right) {\left(\frac{1}{\sqrt{2}}\right)}^{2} + L W = \frac{1}{2} {\left(L + W\right)}^{2}$

because at this point

$\frac{{d}^{2} A}{d {\theta}^{2}} = - 4 \left({W}^{2} + {L}^{2}\right) \cos \theta \sin \theta = - 2 \left({W}^{2} + {L}^{2}\right) < 0$