Geometry Edition: Maximizing Areas and Minimizing Perimeters
Master the art of geometric optimization using calculus and AM-GM inequality for JEE success.
Why Geometric Optimization Matters in JEE
Geometric optimization problems appear in every JEE Main and Advanced paper. These problems test your ability to apply calculus and inequalities to real geometric situations. Mastering this topic can earn you 4-8 guaranteed marks.
🎯 Key Insight
For a given perimeter, the circle has maximum area. For a given area, the circle has minimum perimeter. Among all quadrilaterals with given perimeter, the square has maximum area.
🧭 Problem Navigation
Three Powerful Optimization Methods
📐 Calculus Method
- Formulate function to optimize
- Find derivative
- Set derivative = 0
- Verify maximum/minimum
⚖️ AM-GM Inequality
- For positive numbers
- AM ≥ GM always
- Equality when all equal
- Perfect for fixed sums
📏 Geometric Insight
- Circle for max area/min perimeter
- Square among quadrilaterals
- Equilateral triangle among triangles
- Symmetry principles
Problem 1: Maximum Area Rectangle with Fixed Perimeter
Find the rectangle of maximum area that has a fixed perimeter P.
Solution Using Calculus:
Step 1: Define Variables
Let length = $x$, width = $y$
Perimeter: $2x + 2y = P \Rightarrow y = \frac{P}{2} - x$
Area: $A = xy = x\left(\frac{P}{2} - x\right)$
Step 2: Formulate Function
$$A(x) = \frac{P}{2}x - x^2$$
Domain: $0 < x < \frac{P}{2}$
Step 3: Find Critical Points
$$\frac{dA}{dx} = \frac{P}{2} - 2x$$
Set derivative = 0: $\frac{P}{2} - 2x = 0 \Rightarrow x = \frac{P}{4}$
Step 4: Verify Maximum
$$\frac{d^2A}{dx^2} = -2 < 0$$
Negative second derivative confirms maximum
Step 5: Find Dimensions
$$x = \frac{P}{4}, \quad y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$$
Maximum area occurs when length = width = $\frac{P}{4}$
✅ Maximum Area Rectangle is a SQUARE with side $\frac{P}{4}$
💡 Alternative: AM-GM Method
For fixed sum $x + y = \frac{P}{2}$ (constant), product $xy$ is maximum when $x = y$
By AM-GM: $\frac{x + y}{2} \geq \sqrt{xy} \Rightarrow xy \leq \left(\frac{x + y}{2}\right)^2$
Equality when $x = y = \frac{P}{4}$
Problem 2: Minimum Perimeter Triangle with Fixed Area
Among all triangles with fixed area A, which one has minimum perimeter?
Solution Using AM-GM Inequality:
Step 1: Heron's Formula
Area $A = \sqrt{s(s-a)(s-b)(s-c)}$ where $s = \frac{a+b+c}{2}$
Perimeter $P = 2s$
Step 2: Apply AM-GM
For positive numbers $(s-a), (s-b), (s-c)$:
$$\frac{(s-a)+(s-b)+(s-c)}{3} \geq \sqrt[3]{(s-a)(s-b)(s-c)}$$
But $(s-a)+(s-b)+(s-c) = 3s - (a+b+c) = 3s - 2s = s$
Step 3: Simplify Inequality
$$\frac{s}{3} \geq \sqrt[3]{(s-a)(s-b)(s-c)}$$
Cube both sides: $\frac{s^3}{27} \geq (s-a)(s-b)(s-c)$
Step 4: Relate to Area
From Heron: $A^2 = s(s-a)(s-b)(s-c)$
So $(s-a)(s-b)(s-c) = \frac{A^2}{s}$
Substitute: $\frac{s^3}{27} \geq \frac{A^2}{s}$
Step 5: Find Minimum Perimeter
$$\frac{s^4}{27} \geq A^2 \Rightarrow s^4 \geq 27A^2$$
$$s \geq \sqrt{3\sqrt{3}A}$$
Minimum perimeter $P = 2s = 2\sqrt{3\sqrt{3}A}$
Step 6: Equality Condition
Equality in AM-GM occurs when $s-a = s-b = s-c$
This implies $a = b = c$
✅ Minimum Perimeter Triangle is EQUILATERAL with side $\frac{2}{\sqrt{3}}\sqrt{A\sqrt{3}}$
Problem 3: Maximum Volume Cylinder in a Sphere
Find the right circular cylinder of maximum volume that can be inscribed in a sphere of radius R.
Solution Using Calculus:
Step 1: Define Variables
Let cylinder radius = $r$, height = $h$
From sphere geometry: $r^2 + \left(\frac{h}{2}\right)^2 = R^2$
So $r^2 = R^2 - \frac{h^2}{4}$
Step 2: Volume Function
Cylinder volume: $V = \pi r^2 h = \pi\left(R^2 - \frac{h^2}{4}\right)h$
$$V(h) = \pi R^2 h - \frac{\pi}{4}h^3$$
Domain: $0 < h < 2R$
Step 3: Find Critical Points
$$\frac{dV}{dh} = \pi R^2 - \frac{3\pi}{4}h^2$$
Set derivative = 0: $\pi R^2 - \frac{3\pi}{4}h^2 = 0$
$$h^2 = \frac{4R^2}{3} \Rightarrow h = \frac{2R}{\sqrt{3}}$$
Step 4: Find Radius
$$r^2 = R^2 - \frac{h^2}{4} = R^2 - \frac{R^2}{3} = \frac{2R^2}{3}$$
$$r = R\sqrt{\frac{2}{3}}$$
Step 5: Maximum Volume
$$V_{max} = \pi r^2 h = \pi\left(\frac{2R^2}{3}\right)\left(\frac{2R}{\sqrt{3}}\right) = \frac{4\pi R^3}{3\sqrt{3}}$$
✅ Maximum Volume Cylinder has height $\frac{2R}{\sqrt{3}}$ and radius $R\sqrt{\frac{2}{3}}$
📋 Quick Optimization Formulas
Maximum Area Problems
- Fixed Perimeter: Circle > Square > Rectangle
- Rectangle: Square with side $P/4$
- Triangle: Equilateral triangle
- n-gon: Regular n-gon
Minimum Perimeter Problems
- Fixed Area: Circle < Square < Rectangle
- Rectangle: Square with side $\sqrt{A}$
- Triangle: Equilateral triangle
- General: Most symmetric shape
🎯 Golden Rule: For fixed perimeter, the most symmetrical shape has maximum area. For fixed area, the most symmetrical shape has minimum perimeter.
Problem 4: Optimal Window Design
A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter is fixed as P, find the dimensions for maximum area.
Solution Using Calculus:
Step 1: Define Variables
Let rectangle width = $2r$ (also diameter of semicircle)
Let rectangle height = $h$
Semicircle radius = $r$
Step 2: Perimeter Constraint
Perimeter = Rectangle sides + Semicircle arc
$P = 2h + 2r + \pi r$ (two vertical sides + bottom + semicircle)
So $h = \frac{P - 2r - \pi r}{2} = \frac{P - r(2 + \pi)}{2}$
Step 3: Area Function
Area = Rectangle area + Semicircle area
$A = 2rh + \frac{1}{2}\pi r^2$
Substitute h: $A(r) = 2r\left[\frac{P - r(2 + \pi)}{2}\right] + \frac{1}{2}\pi r^2$
$A(r) = Pr - (2 + \pi)r^2 + \frac{\pi}{2}r^2 = Pr - \left(2 + \frac{\pi}{2}\right)r^2$
Step 4: Optimize
$\frac{dA}{dr} = P - 2\left(2 + \frac{\pi}{2}\right)r = P - (4 + \pi)r$
Set derivative = 0: $P - (4 + \pi)r = 0 \Rightarrow r = \frac{P}{4 + \pi}$
Step 5: Find Height
$h = \frac{P - \frac{P}{4 + \pi}(2 + \pi)}{2} = \frac{P\left(1 - \frac{2 + \pi}{4 + \pi}\right)}{2}$
$h = \frac{P}{2}\left(\frac{4 + \pi - 2 - \pi}{4 + \pi}\right) = \frac{P}{2}\left(\frac{2}{4 + \pi}\right) = \frac{P}{4 + \pi}$
✅ Optimal dimensions: $h = r = \frac{P}{4 + \pi}$, width = $2r = \frac{2P}{4 + \pi}$
Interesting: The rectangle becomes a square when the semicircle is added!
Problem 5: Minimum Fencing Cost
A rectangular field is to be fenced along a straight river (no fencing needed). If fencing costs ₹10/m and the area must be 1800 m², find dimensions for minimum cost.
Solution Using AM-GM:
Step 1: Define Variables
Let length parallel to river = $x$
Let width perpendicular to river = $y$
Area: $xy = 1800 \Rightarrow y = \frac{1800}{x}$
Step 2: Fencing Cost
Fencing needed: $x + 2y$ (one length + two widths)
Cost $C = 10(x + 2y) = 10x + 20y$
Substitute y: $C(x) = 10x + 20\left(\frac{1800}{x}\right) = 10x + \frac{36000}{x}$
Step 3: Apply AM-GM
We need to minimize $10x + \frac{36000}{x}$ for $x > 0$
By AM-GM: $\frac{10x + \frac{36000}{x}}{2} \geq \sqrt{10x \cdot \frac{36000}{x}}$
$\frac{C}{2} \geq \sqrt{360000} = 600$
So $C \geq 1200$
Step 4: Equality Condition
Equality in AM-GM when $10x = \frac{36000}{x}$
$10x^2 = 36000 \Rightarrow x^2 = 3600 \Rightarrow x = 60$
Then $y = \frac{1800}{60} = 30$
Step 5: Verify
Dimensions: $x = 60$ m, $y = 30$ m
Fencing: $60 + 2(30) = 120$ m
Minimum cost: $120 \times 10 = ₹1200$
✅ Minimum cost dimensions: 60 m parallel to river, 30 m perpendicular
🎯 JEE Problem Solving Strategy
When to Use Calculus
- Complex geometric constraints
- Multiple variables involved
- When AM-GM doesn't apply directly
- When you need exact numerical values
- For verification of results
When to Use AM-GM
- Sum of variables is constant
- Product needs to be maximized
- All variables are positive
- Quick mental calculations needed
- For elegant, conceptual solutions
Pro Tip: In JEE, always try AM-GM first - it's faster and more elegant. Use calculus when AM-GM doesn't work or for verification.
📝 Test Your Understanding
Try these optimization problems using the methods you've learned:
1. Find the maximum area of a right triangle with hypotenuse 10 cm.
2. A closed cylindrical can must contain 1 liter. Find dimensions for minimum surface area.
3. Find the rectangle of maximum area that can be inscribed in a circle of radius R.
Mastered Geometric Optimization?
You're now equipped to tackle any JEE optimization problem with confidence!