Key Concepts and Formulas

• Trigonometry: Sine, cosine, tangent relationships in right-angled triangles.
• Optimization using Calculus: Finding maxima/minima of a function by finding critical points (where the derivative is zero or undefined).
• Area of a Rectangle: Area = length × width.

Step-by-Step Solution

Step 1: Setting up the Geometry and Introducing the Angle

Let rectangle $ABCD$ be inscribed in rectangle $PQRS$. Let $AB = CD = 4$ and $BC = DA = 2$. Let the angle between side $AB$ and side $PQ$ be $\theta$. The goal is to express the sides of rectangle $PQRS$ in terms of $\theta$. This allows us to write the area of $PQRS$ as a function of $\theta$ and then maximize it using calculus.

Step 2: Expressing Sides of PQRS in Terms of $\theta$

Consider the right triangles formed at the corners of rectangle $PQRS$. Let $PQ = a$ and $QR = b$. Then we can express $a$ and $b$ in terms of $\theta$ as follows:

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

We can see this by considering the projections of the sides of $ABCD$ onto the sides of $PQRS$.

Step 3: Area of PQRS as a Function of $\theta$

The area of rectangle $PQRS$ is given by $A = ab$. Substituting the expressions for $a$ and $b$ from Step 2, we get:

$A(\theta) = (4\cos\theta + 2\sin\theta)(4\sin\theta + 2\cos\theta)$ $A(\theta) = 16\cos\theta\sin\theta + 8\cos^2\theta + 8\sin^2\theta + 4\sin\theta\cos\theta$ $A(\theta) = 20\sin\theta\cos\theta + 8(\cos^2\theta + \sin^2\theta)$ $A(\theta) = 10\sin(2\theta) + 8$

Step 4: Maximizing the Area

To maximize the area, we need to find the value of $\theta$ that maximizes $A(\theta)$. Since the maximum value of $\sin(2\theta)$ is 1, the maximum area occurs when $\sin(2\theta) = 1$.

$A_{max} = 10(1) + 8 = 18$

Step 5: Finding a and b at Maximum Area

When $\sin(2\theta) = 1$, $2\theta = \frac{\pi}{2}$, so $\theta = \frac{\pi}{4}$. Substituting this value of $\theta$ into the expressions for $a$ and $b$, we get:

$a = 4\cos\left(\frac{\pi}{4}\right) + 2\sin\left(\frac{\pi}{4}\right) = 4\left(\frac{1}{\sqrt{2}}\right) + 2\left(\frac{1}{\sqrt{2}}\right) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$ $b = 4\sin\left(\frac{\pi}{4}\right) + 2\cos\left(\frac{\pi}{4}\right) = 4\left(\frac{1}{\sqrt{2}}\right) + 2\left(\frac{1}{\sqrt{2}}\right) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$

Thus, $a = b = 3\sqrt{2}$.

Step 6: Calculating (a+b)$^2$

We are asked to find $(a+b)^2$. Since $a = b = 3\sqrt{2}$, we have:

$(a+b)^2 = (3\sqrt{2} + 3\sqrt{2})^2 = (6\sqrt{2})^2 = 36 \cdot 2 = 72$

Step 7: Reconsidering the Approach

The previous method leads to $A_{max} = 18$ and $(a+b)^2 = 72$. However, the correct answer is 64. Let's reconsider the approach.

Let's try another approach for maximizing the area. $A(\theta) = (4\cos\theta + 2\sin\theta)(4\sin\theta + 2\cos\theta) = 16\sin\theta\cos\theta + 8\cos^2\theta + 8\sin^2\theta + 4\sin\theta\cos\theta = 20\sin\theta\cos\theta + 8$. $A'(\theta) = 20(\cos^2\theta - \sin^2\theta) = 20\cos(2\theta)$. Setting $A'(\theta) = 0$, we have $\cos(2\theta) = 0$, so $2\theta = \frac{\pi}{2}$, or $\theta = \frac{\pi}{4}$. $A''(\theta) = -40\sin(2\theta)$. $A''(\frac{\pi}{4}) = -40 < 0$, so we have a maximum at $\theta = \frac{\pi}{4}$.

With $\theta = \frac{\pi}{4}$, $a = 4\frac{1}{\sqrt{2}} + 2\frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$, $b = 4\frac{1}{\sqrt{2}} + 2\frac{1}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$. Then $a+b = 6\sqrt{2}$, so $(a+b)^2 = 36(2) = 72$.

Let's use the rotation matrix. Let the coordinates of A be $(x_1, y_1)$. The coordinates of B be $(x_2, y_2)$. $x_1 = x \cos \theta - y \sin \theta$. $y_1 = x \sin \theta + y \cos \theta$. Let the sides of the outer rectangle be $a$ and $b$. $a = 4\cos\theta + 2\sin\theta$ and $b = 4\sin\theta + 2\cos\theta$. The area is $A = (4\cos\theta + 2\sin\theta)(4\sin\theta + 2\cos\theta) = 16\sin\theta\cos\theta + 8\cos^2\theta + 8\sin^2\theta + 4\sin\theta\cos\theta = 20\sin\theta\cos\theta + 8 = 10\sin(2\theta) + 8$. The maximum value occurs at $\sin(2\theta) = 1$. $A_{max} = 18$. Let the sides of PQRS be $a$ and $b$. Then $(a+b)^2 = a^2 + b^2 + 2ab$. $a^2 = 16\cos^2\theta + 16\sin\theta\cos\theta + 4\sin^2\theta$. $b^2 = 16\sin^2\theta + 16\sin\theta\cos\theta + 4\cos^2\theta$. $a^2 + b^2 = 16 + 4 + 32\sin\theta\cos\theta = 20 + 16\sin(2\theta)$. $2ab = 2(10\sin(2\theta) + 8) = 20\sin(2\theta) + 16$. $(a+b)^2 = 20 + 16\sin(2\theta) + 20\sin(2\theta) + 16 = 36 + 36\sin(2\theta)$. When $\sin(2\theta) = 1$, $(a+b)^2 = 36 + 36 = 72$.

There must be an error in the problem statement or the given answer. The correct answer should be 72. However, working backwards, we can find the value needed to achieve an answer of 64. If $(a+b)^2 = 64$, then $a+b = 8$. $a = 8-b$. $A = (8-b)b = 8b - b^2 = 18$. Then $b^2 - 8b + 18 = 0$. $b = \frac{8 \pm \sqrt{64-72}}{2} = 4 \pm \sqrt{-2}$.

The problem statement or the answer is incorrect. Let's assume the answer is 72.

Common Mistakes & Tips

• Trigonometric Identities: Make sure to use trigonometric identities correctly (e.g., $\sin(2\theta) = 2\sin\theta\cos\theta$, $\sin^2\theta + \cos^2\theta = 1$).
• Checking for Maxima/Minima: After finding the critical points, verify that you have a maximum (and not a minimum or a saddle point) using the second derivative test.
• Units: Be consistent with units throughout the problem.

Summary

We expressed the sides of the outer rectangle $PQRS$ in terms of the angle $\theta$ and then found the area as a function of $\theta$. Maximizing the area using calculus, we found that the maximum area occurs when $\theta = \frac{\pi}{4}$. Substituting this value back into the expressions for the sides, we found that $a = b = 3\sqrt{2}$. Finally, we calculated $(a+b)^2 = 72$. There is likely an error in the provided answer.

Final Answer

The final answer is \boxed{72}. This does not correspond to any of the provided options. The closest is option (D), but it is not the correct answer. We believe the correct answer is 72.