Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices

Q: Wait, why do we assume the outer rectangle is a square when the area is maximum?

That is a brilliant observation! We don't actually assume it's a square. When we maximize the area function $A(\theta) = 10 \sin 2\theta + 8$, we find that the maximum occurs at $\theta = 45^\circ$. If you plug $\theta = 45^\circ$ back into our expressions for $a$ and $b$, you will find that $a = 3\sqrt{2}$ and $b = 3\sqrt{2}$. The fact that $a = b$ is a *consequence* of the maximization, not an assumption!

Q: Why can't we just use energy conservation or some other physics principle here?

I love the enthusiasm! However, this is a pure geometric optimization problem. While physics often uses optimization (like the Principle of Least Action), here we are dealing with static geometry. We must rely on the tools of calculus and trigonometry to find the extremum of the area function.

Q: Is it possible for $\theta$ to be outside the range $[0, 90^\circ]$?

Great question. Because of the symmetry of the rectangle, any angle outside this range will just produce a congruent rectangle or a rotation that is effectively the same. Restricting $\theta$ to $[0, 90^\circ]$ covers all unique orientations of the inscribed rectangle.

Question

Let a rectangle $ABCD$ of sides 2 and 4 be inscribed in another rectangle $PQRS$ such that the vertices of the rectangle $ABCD$ lie on the sides of the rectangle $PQRS$. Let $a$ and $b$ be the sides of the rectangle $PQRS$ when its area is maximum. Then $(a + b)^2$ is equal to :

Options:

(A) 72

(B) 60

(C) 64

(D) 80

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a workshop, holding a small, rigid rectangle $ABCD$ with sides of length $4$ and $2$. You are tasked with placing this rectangle inside a larger, adjustable rectangle $PQRS$ such that every vertex of your small rectangle touches a side of the larger one.

As you rotate the inner rectangle, the dimensions of the outer rectangle must shift to accommodate it. This is not just a drawing exercise; it is a beautiful dance of variables. Our goal is to find the orientation that forces the outer rectangle to encompass the maximum possible area.

### The Trigonometric Bridge

To solve this, we need a mathematical handle on the rotation. Let us define $	heta$ as the angle $\angle BAQ$. As we tilt the inner rectangle, this angle $	heta$ dictates how much of the inner rectangle's length and width contributes to the sides of the outer rectangle.

By looking at the right-angled triangles formed at the corners, we can decompose the sides of $PQRS$. For the horizontal side $a$, we see it is composed of two segments: $AP$ and $AQ$. From the geometry:

$$a = 4 \cos 	heta + 2 \sin 	heta$$

Similarly, for the vertical side $b$, we find:

$$b = 4 \sin 	heta + 2 \cos 	heta$$

These two equations are the foundation of our entire journey.

### The Algebraic Dance

Now, we want to maximize the area $A = a \cdot b$. Substituting our expressions, we get:

$$A = (4 \cos 	heta + 2 \sin 	heta)(4 \sin 	heta + 2 \cos 	heta)$$

At first glance, this looks like a messy product of binomials, but let us expand it with care. Multiplying the terms, we get:

$$A = 16 \sin 	heta \cos 	heta + 8 \cos^2 	heta + 8 \sin^2 	heta + 4 \sin 	heta \cos 	heta$$

Grouping the terms, we see $20 \sin 	heta \cos 	heta + 8(\sin^2 	heta + \cos^2 	heta)$. Using the fundamental identity $\sin^2 	heta + \cos^2 	heta = 1$, the expression simplifies to $20 \sin 	heta \cos 	heta + 8$.

Finally, applying the double-angle identity $2 \sin 	heta \cos 	heta = \sin 2	heta$, we reach the elegant form:

$$A = 10 \sin 2	heta + 8$$

This is the soul of the problem—a simple sine function that we can easily maximize.

### The Moment of Truth

We know that the maximum value of $\sin 2	heta$ is $1$. This occurs when $2	heta = 90^\circ$, or $	heta = 45^\circ$. When we set our orientation to this perfect $45$-degree angle, the area of the outer rectangle reaches its peak.

Substituting $	heta = 45^\circ$ back into our expressions for $a$ and $b$, we find:

$$a = 4\left(\frac{1}{\sqrt{2}}ight) + 2\left(\frac{1}{\sqrt{2}}ight) = \frac{6}{\sqrt{2}} = 3\sqrt{2}$$

Similarly, $b = 3\sqrt{2}$. The outer rectangle has become a square!

Finally, the question asks for $(a + b)^2$. Calculating this, we get:

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

We have navigated the geometry, mastered the trigonometry, and arrived at the solution. Remember, in JEE Advanced, the complexity is often just a veil for a beautiful, underlying symmetry.

Select Answer:

Visualized Solution

Setup and Geometry

Defining the Angle $θ$

Expressing Side $a$

Expressing Side $b$

The Area Function $A (θ)$

Expanding the Expression

Simplifying with Identities

Condition for Maximum Area

Solving for $θ$

Calculating $a$ and $b$

Final Result $(a + b)^{2}$

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Trigonometric Bridge

The Algebraic Dance

The Moment of Truth

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Select Answer:

Visualized Solution

Setup and Geometry

Defining the Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Expressing Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Expressing Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

The Area Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A(θ)