The Hook: Maximizing Profit, Minimizing Cost
Harpreet Singh Bedi starts a company and faces a classic problem:
Price too high → Few customers → Low profit Price too low → Many customers but little margin → Low profit
Question: What price maximizes profit?
This is an optimization problem — finding the maximum or minimum value of a function.
If profit $P(x) = -2x^2 + 40x - 50$ (where $x$ is price), we need to find when $P(x)$ is maximum.
Solution: Find when $P'(x) = 0$
$P'(x) = -4x + 40 = 0 \Rightarrow x = 10$
Maximum profit at price = ₹10!
This is the power of maxima and minima — optimization is everywhere in real life.
Interactive: Explore Function Extrema
See how functions reach their highest and lowest points!
What are Maxima and Minima?
Definitions
Local Maximum at $x = a$: $f(a) \geq f(x)$ for all $x$ in a small neighborhood around $a$
Local Minimum at $x = a$: $f(a) \leq f(x)$ for all $x$ in a small neighborhood around $a$
Global (Absolute) Maximum: $f(a) \geq f(x)$ for ALL $x$ in the domain
Global (Absolute) Minimum: $f(a) \leq f(x)$ for ALL $x$ in the domain
Geometric Interpretation
- Local maximum: Peak of a hill (but there might be higher hills elsewhere)
- Global maximum: The highest point on the entire graph
- Local minimum: Bottom of a valley (but there might be deeper valleys)
- Global minimum: The lowest point on the entire graph
Critical Points and Stationary Points
Critical Point: A point $x = a$ where either:
- $f'(a) = 0$ (horizontal tangent), OR
- $f'(a)$ does not exist (sharp corner, vertical tangent, discontinuity)
Stationary Point: A point where $f'(a) = 0$ (subset of critical points)
Key Fact: Maxima and minima can only occur at:
- Critical points (interior of domain)
- Endpoints of the domain
Finding Maxima and Minima: The Process
Step-by-Step Method
To find maxima/minima of $f(x)$ on $[a, b]$:
Step 1: Find $f'(x)$
Step 2: Solve $f'(x) = 0$ to find stationary points
Step 3: Check points where $f'(x)$ doesn’t exist (if any)
Step 4: Determine nature using:
- First Derivative Test, OR
- Second Derivative Test
Step 5: Evaluate $f(x)$ at critical points and endpoints
Step 6: Compare values to find global maximum/minimum
First Derivative Test
First Derivative Test:
Let $x = a$ be a critical point where $f'(a) = 0$.
| Condition | Nature of $x = a$ |
|---|---|
| $f'(x)$ changes from $+$ to $-$ as $x$ increases through $a$ | Local Maximum |
| $f'(x)$ changes from $-$ to $+$ as $x$ increases through $a$ | Local Minimum |
| $f'(x)$ doesn’t change sign | Neither (point of inflection) |
Visual: Think of climbing a mountain:
- Slope goes from uphill (+) to downhill (-) → You’re at the top (maximum)
- Slope goes from downhill (-) to uphill (+) → You’re at the bottom (minimum)
Example 1
Find local extrema of $f(x) = x^3 - 3x^2 - 9x + 5$
Step 1: $f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$
Step 2: $f'(x) = 0 \Rightarrow x = -1, 3$
Step 3: Sign analysis
| Interval | $x < -1$ | $x = -1$ | $-1 < x < 3$ | $x = 3$ | $x > 3$ |
|---|---|---|---|---|---|
| $f'(x)$ | $+$ | $0$ | $-$ | $0$ | $+$ |
| $f(x)$ | Increasing | Max | Decreasing | Min | Increasing |
Conclusion:
- $x = -1$: Local maximum (slope changes from + to -)
- $x = 3$: Local minimum (slope changes from - to +)
Second Derivative Test
Second Derivative Test:
Let $f'(a) = 0$ and $f''(a)$ exists. Then:
| Condition | Nature of $x = a$ |
|---|---|
| $f''(a) > 0$ | Local Minimum (concave up ∪) |
| $f''(a) < 0$ | Local Maximum (concave down ∩) |
| $f''(a) = 0$ | Test fails (use first derivative test) |
Intuition:
- $f''(a) > 0$: Curve bends upward → bottom of a cup → minimum
- $f''(a) < 0$: Curve bends downward → top of a hill → maximum
Example 2
Find extrema of $f(x) = x^4 - 4x^3$
Step 1: $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$
Step 2: $f'(x) = 0 \Rightarrow x = 0, 3$
Step 3: Second derivative test
$f''(x) = 12x^2 - 24x = 12x(x - 2)$
At $x = 0$: $f''(0) = 0$ (test fails)
At $x = 3$: $f''(3) = 12(3)(1) = 36 > 0$ → Local minimum
For $x = 0$, use first derivative test:
- For $x < 0$: $f'(x) = 4x^2(x-3) < 0$ (negative × negative = positive? NO, $(x-3)$ is negative)
- Actually: $x = -0.1$: $f'(-0.1) = 4(0.01)(-3.1) < 0$
- $x = 0.1$: $f'(0.1) = 4(0.01)(-2.9) < 0$
Slope doesn’t change sign → $x = 0$ is neither max nor min (inflection point)
Values:
- $f(0) = 0$
- $f(3) = 81 - 108 = -27$ ← Local minimum
Special Cases and Important Results
Case 1: Points Where Derivative Doesn’t Exist
Example: $f(x) = |x|$ on $[-2, 2]$
$f'(x)$ doesn’t exist at $x = 0$ (sharp corner)
But $f(0) = 0$ is the global minimum!
Lesson: Always check points where $f'$ doesn’t exist.
Case 2: Endpoints of Closed Interval
Example: $f(x) = x^2$ on $[1, 3]$
$f'(x) = 2x = 0 \Rightarrow x = 0$ (not in $[1, 3]$)
Check endpoints:
- $f(1) = 1$ ← Global minimum
- $f(3) = 9$ ← Global maximum
Lesson: Always evaluate at endpoints!
Case 3: No Global Extrema
Example: $f(x) = x^3$ on $(-\infty, \infty)$
$f'(x) = 3x^2 = 0 \Rightarrow x = 0$
But $f''(0) = 0$ and first derivative test shows $f'(x) \geq 0$ everywhere (always increasing)
$x = 0$ is neither max nor min (inflection point)
No global max or min exists!
Optimization Problems: Real-World Applications
Template for Word Problems
Optimization Problem Steps:
- Read carefully — Identify what to maximize/minimize
- Draw a diagram if geometric
- Define variables — Let $x$ be the unknown
- Write the objective function — Express what you want to optimize in terms of $x$
- Find constraints — Any restrictions on $x$?
- Differentiate — Find $f'(x) = 0$
- Check nature — Use second derivative or first derivative test
- Answer the question — Don’t forget units!
Example 1: Maximum Area Problem
Problem: A farmer has 100 m of fencing. What dimensions maximize the area of a rectangular enclosure?
Solution:
Let length = $x$, width = $y$
Constraint: Perimeter = $2x + 2y = 100 \Rightarrow y = 50 - x$
Objective: Maximize area $A = xy = x(50 - x) = 50x - x^2$
Differentiate: $\frac{dA}{dx} = 50 - 2x = 0 \Rightarrow x = 25$
Second derivative: $\frac{d^2A}{dx^2} = -2 < 0$ → Maximum ✓
Answer: Length = 25 m, Width = 25 m (square gives maximum area)
Example 2: Minimum Distance Problem
Problem: Find the point on the curve $y = x^2$ closest to $(0, 1)$.
Solution:
Point on curve: $(x, x^2)$
Distance: $D = \sqrt{(x-0)^2 + (x^2-1)^2} = \sqrt{x^2 + (x^2-1)^2}$
Tip: Minimize $D^2$ instead (easier, same result):
$$f(x) = D^2 = x^2 + (x^2 - 1)^2 = x^2 + x^4 - 2x^2 + 1 = x^4 - x^2 + 1$$Differentiate: $f'(x) = 4x^3 - 2x = 2x(2x^2 - 1) = 0$
$x = 0$ or $x^2 = \frac{1}{2} \Rightarrow x = \pm\frac{1}{\sqrt{2}}$
Second derivative: $f''(x) = 12x^2 - 2$
- $f''(0) = -2 < 0$ → Maximum (furthest point)
- $f''\left(\frac{1}{\sqrt{2}}\right) = 12 \cdot \frac{1}{2} - 2 = 4 > 0$ → Minimum ✓
Answer: Closest points are $\left(\pm\frac{1}{\sqrt{2}}, \frac{1}{2}\right)$
Example 3: Minimum Cost Problem (JEE Type)
Problem: A box with square base and open top must have volume 32 m³. Find dimensions that minimize material used.
Solution:
Let base side = $x$, height = $h$
Constraint: Volume = $x^2h = 32 \Rightarrow h = \frac{32}{x^2}$
Surface area (material): $S = x^2 + 4xh$ (base + 4 sides, no top)
$$S = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}$$Differentiate: $\frac{dS}{dx} = 2x - \frac{128}{x^2} = 0$
$$2x = \frac{128}{x^2} \Rightarrow 2x^3 = 128 \Rightarrow x^3 = 64 \Rightarrow x = 4$$Second derivative: $\frac{d^2S}{dx^2} = 2 + \frac{256}{x^3}$
At $x = 4$: $\frac{d^2S}{dx^2} = 2 + \frac{256}{64} = 2 + 4 = 6 > 0$ → Minimum ✓
Answer: Base = 4 m × 4 m, Height = $\frac{32}{16} = 2$ m
Advanced Concepts
Absolute Maxima/Minima on Closed Interval
Extreme Value Theorem:
If $f(x)$ is continuous on closed interval $[a, b]$, then $f$ attains both absolute maximum and absolute minimum on $[a, b]$.
Method to find them:
- Find all critical points in $(a, b)$
- Evaluate $f$ at critical points and endpoints $a, b$
- Largest value = absolute max, smallest = absolute min
Multiple Critical Points
When a function has many critical points, create a table:
| $x$ | $f(x)$ | Type |
|---|---|---|
| Critical point 1 | Value | Max/Min/Neither |
| Critical point 2 | Value | Max/Min/Neither |
| Endpoint 1 | Value | - |
| Endpoint 2 | Value | - |
Compare all values to determine global extrema.
Memory Tricks
- $f''(x) > 0$: Concave UP (∪) → Like a CUP → Holds water → Minimum
- $f''(x) < 0$: Concave down (∩) → Like a CAP → Water falls off → Maximum
“Positive second derivative = Cup = Minimum”
- + to -: Climbing up then going down → At the top → Maximum
- - to +: Going down then climbing up → At the bottom → Minimum
“Plus-to-minus = Maximum, Minus-to-plus = Minimum”
Define variables Constraint equation Differentiate Check nature (2nd derivative)
“Define, Constrain, Differentiate, Check!”
Practice Problems
Level 1: Foundation (NCERT)
Question: Find local maxima/minima of $f(x) = x^3 - 6x^2 + 9x + 15$
Solution:
$$f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x-1)(x-3)$$Critical points: $x = 1, 3$
Second derivative: $f''(x) = 6x - 12$
At $x = 1$: $f''(1) = -6 < 0$ → Local maximum, $f(1) = 19$
At $x = 3$: $f''(3) = 6 > 0$ → Local minimum, $f(3) = 15$
Question: Find absolute max/min of $f(x) = x^2 - 4x + 6$ on $[0, 4]$
Solution:
$$f'(x) = 2x - 4 = 0 \Rightarrow x = 2$$Evaluate:
- $f(0) = 6$
- $f(2) = 4 - 8 + 6 = 2$ ← Absolute minimum
- $f(4) = 16 - 16 + 6 = 6$
Absolute maximum = 6 (at $x = 0$ and $x = 4$)
Level 2: JEE Main
Question: Divide 10 into two parts such that their product is maximum.
Solution:
Let parts be $x$ and $10 - x$
Product: $P = x(10 - x) = 10x - x^2$
$$P'(x) = 10 - 2x = 0 \Rightarrow x = 5$$ $$P''(x) = -2 < 0$$→ Maximum
Answer: Parts are 5 and 5.
Question: A wire of length 28 cm is bent into a rectangle. What dimensions give maximum area?
Solution:
Let length = $x$, width = $y$
Perimeter: $2x + 2y = 28 \Rightarrow y = 14 - x$
Area: $A = xy = x(14 - x) = 14x - x^2$
$$A'(x) = 14 - 2x = 0 \Rightarrow x = 7$$ $$A''(x) = -2 < 0$$→ Maximum
Answer: 7 cm × 7 cm (square)
Question: Find two positive numbers whose sum is 16 and sum of whose squares is minimum.
Solution:
Let numbers be $x$ and $16 - x$
Sum of squares: $S = x^2 + (16-x)^2 = x^2 + 256 - 32x + x^2 = 2x^2 - 32x + 256$
$$S'(x) = 4x - 32 = 0 \Rightarrow x = 8$$ $$S''(x) = 4 > 0$$→ Minimum
Answer: Numbers are 8 and 8.
Level 3: JEE Advanced
Question: A cylinder is inscribed in a sphere of radius $R$. Find the maximum volume of the cylinder.
Solution:
Let cylinder radius = $r$, height = $h$
By Pythagorean theorem: $r^2 + \left(\frac{h}{2}\right)^2 = R^2$
$$r^2 = R^2 - \frac{h^2}{4}$$Volume: $V = \pi r^2 h = \pi\left(R^2 - \frac{h^2}{4}\right)h = \pi R^2h - \frac{\pi h^3}{4}$
$$\frac{dV}{dh} = \pi R^2 - \frac{3\pi h^2}{4} = 0$$ $$R^2 = \frac{3h^2}{4} \Rightarrow h = \frac{2R}{\sqrt{3}}$$ $$r^2 = R^2 - \frac{h^2}{4} = R^2 - \frac{R^2}{3} = \frac{2R^2}{3}$$Maximum volume: $V = \pi \cdot \frac{2R^2}{3} \cdot \frac{2R}{\sqrt{3}} = \frac{4\pi R^3}{3\sqrt{3}} = \frac{4\pi R^3\sqrt{3}}{9}$
Question: A window has the shape of a rectangle surmounted by a semicircle. If the perimeter is 10 m, find dimensions for maximum light (area).
Solution:
Let rectangle width = $2r$ (diameter of semicircle), height = $h$
Perimeter: $2h + 2r + \pi r = 10$
$$h = \frac{10 - 2r - \pi r}{2} = 5 - r - \frac{\pi r}{2}$$Area: $A = 2rh + \frac{\pi r^2}{2}$
$$A = 2r\left(5 - r - \frac{\pi r}{2}\right) + \frac{\pi r^2}{2}$$ $$= 10r - 2r^2 - \pi r^2 + \frac{\pi r^2}{2} = 10r - 2r^2 - \frac{\pi r^2}{2}$$ $$\frac{dA}{dr} = 10 - 4r - \pi r = 0$$ $$r = \frac{10}{4 + \pi}$$ $$h = 5 - r - \frac{\pi r}{2} = 5 - \frac{10}{4+\pi} - \frac{5\pi}{4+\pi} = \frac{5(4+\pi) - 10 - 5\pi}{4+\pi} = \frac{10}{4+\pi}$$Answer: $r = \frac{10}{4+\pi}$ m, $h = \frac{10}{4+\pi}$ m (height = radius!)
Question: Prove that of all rectangles with given perimeter, the square has maximum area.
Solution:
Let perimeter = $P$, length = $x$, width = $y$
Constraint: $2x + 2y = P \Rightarrow y = \frac{P}{2} - x$
Area: $A = xy = x\left(\frac{P}{2} - x\right) = \frac{Px}{2} - x^2$
$$\frac{dA}{dx} = \frac{P}{2} - 2x = 0 \Rightarrow x = \frac{P}{4}$$ $$y = \frac{P}{2} - \frac{P}{4} = \frac{P}{4}$$Since $x = y$, the rectangle is a square.
$$\frac{d^2A}{dx^2} = -2 < 0$$→ Maximum ∎
Common Mistakes to Avoid
When finding absolute max/min on $[a, b]$, always evaluate at endpoints!
Critical points give local extrema, but global extrema might be at boundaries.
If $f''(a) = 0$, second derivative test fails. Must use first derivative test instead!
Example: $f(x) = x^4$ at $x = 0$: $f'(0) = 0$, $f''(0) = 0$ (test fails), but $x = 0$ is actually a minimum.
In optimization problems, check if your solution makes physical sense!
If you get $x = -5$ meters for a length, something’s wrong!
Problem: “Find dimensions that minimize cost…”
Don’t minimize area/volume — minimize cost (which may depend on area/volume)!
Quick Revision Box
| Test | Condition | Result |
|---|---|---|
| First Derivative | $f'$ changes + to - | Local Max |
| $f'$ changes - to + | Local Min | |
| Second Derivative | $f'(a) = 0$ and $f''(a) > 0$ | Local Min (∪) |
| $f'(a) = 0$ and $f''(a) < 0$ | Local Max (∩) | |
| $f''(a) = 0$ | Test fails |
Critical Points: $f'(x) = 0$ or $f'(x)$ doesn’t exist
Absolute Extrema on $[a, b]$: Check critical points + endpoints
JEE Strategy Tips
Weightage: Maxima-minima is HUGE in JEE — 4-5 questions across Main & Advanced!
Time-Saver: In optimization, if objective function has only one critical point and problem makes sense, you can often skip checking nature (it must be max/min).
Common Pattern: JEE loves:
- Cylinder/cone in sphere problems
- Rectangle/square area/perimeter problems
- Distance minimization problems
Trap Alert: Always check domain restrictions! Negative lengths/areas are meaningless.
Advanced Tip: For complex optimization, sometimes AM-GM inequality is faster than calculus!
Exam Hack: If asked “prove square gives max area” or similar, use second derivative to show it’s truly a maximum, not just $f'(x) = 0$.
Teacher’s Summary
Maxima/minima occur at critical points (where $f'(x) = 0$ or doesn’t exist) or at endpoints.
First derivative test: Check sign change of $f'(x)$ around critical point.
Second derivative test: $f''(a) > 0$ → min, $f''(a) < 0$ → max (faster but can fail if $f'' = 0$).
Optimization strategy: Define variables → Write constraint → Express objective function → Differentiate → Check nature.
Absolute extrema on $[a, b]$: Evaluate at ALL critical points AND endpoints, then compare.
“In life and calculus, finding the maximum and minimum is all about knowing where to look!”
Related Topics
Within Limits, Continuity & Differentiability
- Higher Derivatives — Second derivative test uses $f''(x)$
- Mean Value Theorems — Rolle’s theorem connects to maxima/minima
- Curve Sketching — Using extrema to sketch curves
- Applications of Derivatives — Broader applications
Mathematical Foundations
- Inequalities — AM-GM for optimization
- Coordinate Geometry — Distance problems
Applications
- Area Under Curves — Related optimization
- Differential Equations — Finding extrema of solutions
Physics Connections
- Projectile Motion — Maximum height/range
- Energy Conservation — Minimum energy configurations
- Optics — Lens optimization