1. Set up the figure

Let the inner rectangle $ABCD$ have sides $4$ and $2$. Assume it is rotated by an angle $\theta$ with respect to the outer rectangle $PQRS$.

For a rectangle of width $4$ and height $2$, when rotated by angle $\theta$, the dimensions of the smallest axis-parallel bounding rectangle are:

$a = 4\cos\theta + 2\sin\theta$ $b = 4\sin\theta + 2\cos\theta$

So the area of the outer rectangle is

$A(\theta)=ab=(4\cos\theta+2\sin\theta)(4\sin\theta+2\cos\theta).$

2. Expand the area expression

$A(\theta)=16\sin\theta\cos\theta+8\cos^2\theta+8\sin^2\theta+4\sin\theta\cos\theta$

$A(\theta)=8(\sin^2\theta+\cos^2\theta)+20\sin\theta\cos\theta$

Since $\sin^2\theta+\cos^2\theta=1$,

$A(\theta)=8+20\sin\theta\cos\theta.$

Using

$2\sin\theta\cos\theta=\sin 2\theta,$

we get

$A(\theta)=8+10\sin 2\theta.$

3. Maximize the area

Since $\sin 2\theta \le 1$, the maximum area occurs when

$\sin 2\theta=1 \implies 2\theta=\frac{\pi}{2} \implies \theta=\frac{\pi}{4}.$

So at maximum area,

$\cos\theta=\sin\theta=\frac{1}{\sqrt{2}}.$

Then

$a=4\cdot \frac{1}{\sqrt{2}}+2\cdot \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}=3\sqrt{2},$

and similarly,

$b=4\cdot \frac{1}{\sqrt{2}}+2\cdot \frac{1}{\sqrt{2}}=3\sqrt{2}.$

Thus,

$a+b=6\sqrt{2}.$

Hence,

$(a+b)^2=(6\sqrt{2})^2=72.$

4. Check options

• A: $64$
• B: $80$
• C: $60$
• D: $72$

Therefore, the correct option is:

$\boxed{\text{D}}$