The Rectangle in a Curve: A Classic JEE Problem (and its many variants)
Master optimization in coordinate geometry with this fundamental problem type that appears in various forms across JEE Main and Advanced.
Why This Problem is a JEE Favorite
The "rectangle in a curve" problem beautifully combines coordinate geometry, calculus optimization, and algebraic manipulation - making it perfect for testing multiple mathematical skills in a single question.
📐 Visualizing the Classic Problem
Find rectangle with maximum area that fits under the curve
🎯 JEE Significance
- Appears in JEE Main 2016, 2018, 2021 and JEE Advanced 2015, 2019
- Tests application of derivatives in geometric contexts
- Perfect for multi-concept questions worth 4-8 marks
- Foundation for more complex optimization problems
🧭 Problem Variants Navigation
Variant 1: The Classic Parabola Problem
Find the area of the largest rectangle that can be inscribed in the parabola $y = 12 - x^2$ with its base on the x-axis.
Solution Approach:
Step 1: Define Variables
Let the rectangle have vertices at $(x, 0)$, $(-x, 0)$, $(x, 12-x^2)$, $(-x, 12-x^2)$
Width = $2x$, Height = $12 - x^2$
Step 2: Area Function
Area $A(x) = \text{width} \times \text{height} = 2x(12 - x^2) = 24x - 2x^3$
Domain: $0 \leq x \leq \sqrt{12}$ (since height must be non-negative)
Step 3: Find Critical Points
$A'(x) = 24 - 6x^2 = 0$
$6x^2 = 24 \Rightarrow x^2 = 4 \Rightarrow x = 2$ (positive root)
Step 4: Verify Maximum
$A''(x) = -12x$
$A''(2) = -24 < 0$ ⇒ Maximum at $x = 2$
Step 5: Calculate Maximum Area
$A(2) = 24(2) - 2(2)^3 = 48 - 16 = 32$
Final Answer
Maximum Area = $32$ square units
Dimensions: Width = 4 units, Height = 8 units
💡 Key Insight
For symmetric curves like parabolas centered on y-axis, the maximum area rectangle is always symmetric about the y-axis. This reduces the problem to a single-variable optimization.
Variant 2: Rectangle in an Ellipse
Find the maximum area of a rectangle inscribed in the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.
Solution Approach:
Step 1: Parameterize the Ellipse
Let vertex of rectangle in first quadrant be $(4\cos\theta, 3\sin\theta)$
Due to symmetry, rectangle vertices: $(\pm4\cos\theta, \pm3\sin\theta)$
Step 2: Area Function
Width = $8\cos\theta$, Height = $6\sin\theta$
Area $A(\theta) = 48\cos\theta\sin\theta = 24\sin(2\theta)$
Step 3: Optimization
$A(\theta) = 24\sin(2\theta)$ is maximum when $\sin(2\theta) = 1$
$\Rightarrow 2\theta = \frac{\pi}{2} \Rightarrow \theta = \frac{\pi}{4}$
Step 4: Maximum Area
$A_{\text{max}} = 24\sin\left(2\cdot\frac{\pi}{4}\right) = 24\sin\left(\frac{\pi}{2}\right) = 24$
Final Answer
Maximum Area = $24$ square units
Dimensions: Width = $4\sqrt{2}$, Height = $3\sqrt{2}$
🔄 Related Variant
Rectangle with sides parallel to axes: Let vertices be $(x, y)$ in first quadrant, then area $A = 4xy$ with constraint $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.
Using AM-GM: $\frac{x^2}{a^2} + \frac{y^2}{b^2} \geq 2\frac{xy}{ab} \Rightarrow xy \leq \frac{ab}{2} \Rightarrow A_{\text{max}} = 2ab$
Variant 3: Rectangle in Right Triangle
Find the maximum area of a rectangle inscribed in a right triangle with legs of lengths 6 and 8 units, with one side of rectangle along the hypotenuse.
Solution Approach:
Step 1: Set Up Coordinate System
Place right angle at origin: vertices at (0,0), (6,0), (0,8)
Hypotenuse equation: $\frac{x}{6} + \frac{y}{8} = 1$ or $4x + 3y = 24$
Step 2: Define Rectangle Position
Let rectangle have vertices at (x,0), (6,0), (x,y), (6,y) with (x,y) on hypotenuse
From hypotenuse: $4x + 3y = 24 \Rightarrow y = 8 - \frac{4}{3}x$
Step 3: Area Function
Width = $6 - x$, Height = $y = 8 - \frac{4}{3}x$
Area $A(x) = (6-x)\left(8 - \frac{4}{3}x\right)$
$A(x) = 48 - 8x - 8x + \frac{4}{3}x^2 = 48 - 16x + \frac{4}{3}x^2$
Step 4: Find Critical Point
$A'(x) = -16 + \frac{8}{3}x = 0$
$\frac{8}{3}x = 16 \Rightarrow x = 6$
Step 5: Check Endpoints
$A(0) = 48$, $A(6) = 0$
Maximum at $x = 0$? But this gives degenerate rectangle.
Actually, need to check if vertex lies within triangle boundaries.
Important Note
This variant requires careful boundary analysis. The maximum occurs when the rectangle touches all three sides of the triangle.
🔧 Universal Problem-Solving Framework
Step-by-Step Approach
- Visualize & Set Coordinates: Place the curve in convenient position
- Define Variables: Choose parameters for rectangle position/size
- Area Function: Express area in terms of your variables
- Constraints: Incorporate curve equation as constraint
- Optimize: Use calculus or AM-GM inequality
- Verify: Check boundaries and second derivative
Common Optimization Techniques
- Calculus: Derivatives for critical points
- AM-GM Inequality: For product maximization
- Trigonometric Substitution: For ellipses and circles
- Symmetry Arguments: Reduce variables
- Boundary Analysis: Check edge cases
⚡ Quick Practice Problems
Test your understanding with these variations:
1. Maximum area rectangle in semicircle $y = \sqrt{4 - x^2}$
2. Rectangle under curve $y = e^{-x}$ with one vertex at origin
3. Maximum perimeter rectangle in ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$
🎯 JEE Exam Strategy
Time Management
- 5-7 minutes for standard variants
- Recognize symmetry to reduce variables quickly
- Use AM-GM when possible (faster than calculus)
- If stuck, try specific positions (corners, center)
Common Pitfalls
- Forgetting to check boundary conditions
- Missing symmetry that simplifies the problem
- Incorrect constraint equations
- Not verifying maximum vs minimum
Mastered Rectangle in Curve Problems?
You've learned one of the most reusable problem-solving frameworks in JEE mathematics