If a rectangle is inscribed in an equilateral triangle of side length 2√2 as shown in the figure, th

Q: Why can't we just assume the rectangle's length is half the triangle's side? It looks like it should be symmetric.

That is a very intuitive trap! While the setup is symmetric, the 'optimal' rectangle isn't necessarily half the side length by visual intuition alone. We must rely on the calculus of the area function. The derivative tells us the truth, not our eyes. In this specific case, the math confirms the optimal length is indeed half the side, but you must prove it to be certain.

Q: Can I use AM-GM inequality instead of differentiation?

Absolutely! You are thinking like a topper. Since $A = \frac{\sqrt{3}}{2} l(a-l)$, you can rewrite this as $\frac{\sqrt{3}}{2} [l(a-l)]$. Since the sum of $l$ and $(a-l)$ is constant ($a$), their product is maximized when $l = a-l$, i.e., $l = a/2$. Calculus is the standard path, but AM-GM is the elegant shortcut.

Q: I got the area as $\sqrt{3}$, but the answer is 3. Did I do something wrong?

You did everything perfectly! The question asks for the *square* of the largest area. Always re-read the final line of the question. It is a classic JEE tactic to ask for a transformation of the final result to test your attention to detail.

Question

If a rectangle is inscribed in an equilateral triangle of side length $2\sqrt{2}$ as shown in the figure, then the square of the largest area of such a rectangle is ___

Sigma Percentile · Accepted Answer

### The Geometry of Optimization: A Journey into the Inscribed Rectangle

Welcome, future engineer. Today, we are not just solving a geometry problem; we are embarking on a journey of optimization.

When you look at an equilateral triangle with a rectangle tucked inside, don't just see lines and polygons. See a system in equilibrium. See a challenge that demands you find the 'perfect' balance between length and breadth.

This is the essence of JEE Advanced—finding the peak of a function amidst the constraints of geometry.

### Phase 1: The Visual Anchor

Imagine you are standing before this equilateral triangle. Its side length is given as $a = 2\sqrt{2}$. It is a rigid, unyielding structure.

Now, we place a rectangle inside it. The rectangle's base sits on the triangle's base, and its top two vertices touch the sides of the triangle.

As you adjust the width of this rectangle, its height must change to keep those top corners touching the triangle's edges. This is the 'dance' of the variables. Let the length of the rectangle be $l$ and its breadth be $b$. Our mission is to maximize the area $A = l \cdot b$.

### Phase 2: The Mathematical Bridge

To solve this, we need a bridge between $l$ and $b$. Look at the corners of the triangle. When you place the rectangle, you create two small right-angled triangles at the base.

Because the large triangle is equilateral, its base angles are $60^\circ$. This is our key. The total base of the triangle is $a$.

The rectangle takes up a length $l$ in the middle. By symmetry, the remaining base length is $a - l$, which is split equally between the two small triangles on the sides. Thus, the base of one small right-angled triangle is $\frac{a - l}{2}$.

Now, apply trigonometry. In that small right triangle, the height is $b$ and the base is $\frac{a - l}{2}$. The angle is $60^\circ$. Therefore, we have the relationship:

$$ 	an 60^\circ = \frac{b}{\frac{a - l}{2}} $$

Since $	an 60^\circ = \sqrt{3}$, we can rearrange this to find $b$:

$$ \sqrt{3} = \frac{2b}{a - l} \implies b = \frac{\sqrt{3}}{2}(a - l) $$

This equation is the heart of our problem. It tells us exactly how the height $b$ must shrink as the length $l$ grows. It is a linear constraint, a beautiful trade-off.

### Phase 3: The Calculus of Optimization

Now, let us construct our Area function. We know $A = l \cdot b$. Substituting our expression for $b$, we get:

$$ A(l) = l \cdot \frac{\sqrt{3}}{2}(a - l) = \frac{\sqrt{3}}{2}(al - l^2) $$

This is a downward-opening parabola. We know from our study of functions that the maximum of a parabola occurs at its vertex. To find this, we take the derivative with respect to $l$ and set it to zero:

$$ \frac{dA}{dl} = \frac{\sqrt{3}}{2}(a - 2l) = 0 $$

Solving this, we find the optimal length:

$$ l = \frac{a}{2} $$

This is a profound result. It tells us that for any equilateral triangle, the rectangle of maximum area will always have a length exactly half the side of the triangle. It is elegant, simple, and powerful.

### Phase 4: The Final Calculation

We are given $a = 2\sqrt{2}$. Substituting this into our optimal length:

$$ l = \frac{2\sqrt{2}}{2} = \sqrt{2} $$

Now, find the corresponding breadth $b$:

$$ b = \frac{\sqrt{3}}{2}(2\sqrt{2} - \sqrt{2}) = \frac{\sqrt{3}}{2}(\sqrt{2}) = \sqrt{\frac{3}{2}} $$

Finally, calculate the maximum area $A$:

$$ A = l \cdot b = \sqrt{2} \cdot \sqrt{\frac{3}{2}} = \sqrt{3} $$

But wait! Do not stop here. The question asks for the *square* of the largest area. We must square our result:

$$ A^2 = (\sqrt{3})^2 = 3 $$

### Conclusion

There it is. The answer is **3**.

You have navigated the geometry, set up the trigonometric relationship, performed the calculus, and arrived at the solution. This is how you conquer JEE problems—not by memorizing formulas, but by understanding the geometric soul of the question. Keep this clarity of thought, and you will be unstoppable.

Animated Solution for Mathematics - Differentiation: If a rectangle is inscribed in an equilateral triangle of side length $22$ as shown in the figure, then the square of the largest area of such a rectangle is ___

Enter Numerical Value:

Visualized Solution

Visual Anchor: Inscribed Rectangle

Define Dimensions

Geometric Symmetry

Trigonometric Relation

Express $b$ in terms of $l$

Area Function Setup

Condition for Maxima

Solve for Optimal Length $l$

Calculate Optimal Breadth $b$

Calculate Maximum Area $A$

Final Result

The Sigma Insight: Maxima and Minima

The Geometry of Optimization

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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