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Maxima, Minima & Optimisation

Master Maxima Minima and Optimisation for JEE Main JEE Advanced and Board Exams. Covers local and global extrema first and second derivative tests higher order test closed interval method practical optimisation problems and AM-GM method with solved JEE questions.

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Finding the highest and lowest values of a function — maxima and minima — is perhaps the single most practically important application of calculus. From designing the most efficient container to finding the shortest path, from minimising cost to maximising profit, optimisation problems pervade science, engineering, and economics. The derivative provides the precise mathematical tool to locate these extreme values. For JEE (Main & Advanced), maxima-minima is one of the highest-weightage sub-topics in calculus — expect 2–3 questions in every paper, ranging from straightforward critical-point problems to intricate JEE Advanced optimisation challenges.

1. Local (Relative) Maxima and Minima

A function $f$ has a local maximum at $x = c$ if $f (c) \geq f (x)$ for all $x$ in some neighbourhood of $c$ , and a local minimum if $f (c) \leq f (x)$ for all $x$ near $c$ .

A point $x = c$ where $f^{'} (c) = 0$ or $f^{'} (c)$ does not exist is called a critical point. Every local extremum occurs at a critical point, but not every critical point is an extremum — it could be a point of inflection.

First Derivative Test

At a critical point $x = c$ where $f^{'} (c) = 0$ :

Sign of $f^{'} (x)$ just left of $c$	Sign of $f^{'} (x)$ just right of $c$	Nature of $c$
$+$ (positive)	$-$ (negative)	Local Maximum
$-$ (negative)	$+$ (positive)	Local Minimum
$+$ (positive)	$+$ (positive)	Point of Inflection (neither)
$-$ (negative)	$-$ (negative)	Point of Inflection (neither)

Second Derivative Test

At a critical point $c$ where $f^{'} (c) = 0$ :

If $f^{″} (c) < 0$ : Local Maximum (concave downward — curve is like $\cap$ ).
If $f^{″} (c) > 0$ : Local Minimum (concave upward — curve is like $\cup$ ).
If $f^{″} (c) = 0$ : Test inconclusive — use first derivative test or higher derivative test.

Higher Order Derivative Test (when $f^{″} (c) = 0$ )

Find the first non-zero derivative at $c$ . If $f^{(n)} (c) \neq 0$ and $f^{'} (c) = f^{″} (c) = \dots = f^{(n - 1)} (c) = 0$ :

$n$ is odd: point of inflection (regardless of sign).
$n$ is even and $f^{(n)} (c) > 0$ : local minimum.
$n$ is even and $f^{(n)} (c) < 0$ : local maximum.

Example: $f (x) = x^{4}$ : $f^{'} (0) = f^{″} (0) = f^{‴} (0) = 0$ , $f^{(4)} (0) = 24 > 0$ , $n = 4$ (even, positive) → local minimum at $x = 0$ ✓.

2. Absolute (Global) Maxima and Minima on $[a, b]$

The Closed Interval Method finds the absolute maximum and minimum of a continuous function $f$ on $[a, b]$ :

Step 1: Find all critical points in the open interval $(a, b)$ — solve $f^{'} (x) = 0$ and find where $f^{'} (x)$ is undefined.
Step 2: Evaluate $f$ at all critical points AND at both endpoints $x = a$ and $x = b$ .
Step 3: The largest of all these values is the absolute maximum; the smallest is the absolute minimum.

The absolute extremum might occur at a critical point or at an endpoint — never assume it must be at a critical point.

Worked Example

Find the absolute maximum and minimum of $f (x) = x^{3} - 3 x^{2} + 1$ on $[- \frac{1}{2}, 4]$ .

$f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2)$ .
Critical points in $(- \frac{1}{2}, 4)$ : $x = 0$ and $x = 2$ .
Values: $f (- \frac{1}{2}) = - \frac{1}{8} - \frac{3}{4} + 1 = \frac{1}{8}$ ;
$f (0) = 1$ ; $f (2) = 8 - 12 + 1 = - 3$ ; $f (4) = 64 - 48 + 1 = 17$ .
Absolute maximum: $17$ at $x = 4$ ; Absolute minimum: $- 3$ at $x = 2$ .

3. Optimisation — Practical Maxima/Minima Problems

Optimisation problems ask for the maximum or minimum of a real-world quantity subject to a constraint. Standard approach:

Step 1: Define variables and identify the objective function (what to maximise/minimise) and the constraint.
Step 2: Use the constraint to express the objective function in terms of a single variable.
Step 3: Differentiate and set $= 0$ to find critical points. State the domain of the variable.
Step 4: Verify maximum/minimum using the second derivative test or first derivative test.
Step 5: Check boundary values if the variable has a restricted domain.

Classic Optimisation Results to Memorise

Problem	Optimal Result
Rectangle with fixed perimeter — max area	Square has maximum area
Rectangle with fixed area — min perimeter	Square has minimum perimeter
Cylinder of max volume inscribed in sphere of radius $R$	Height $= \frac{2 R}{\sqrt{3}}$ , radius $= R \sqrt{\frac{2}{3}}$
Cone of max volume inscribed in sphere of radius $R$	Height $= \frac{4 R}{3}$
Largest rectangle inscribed in semicircle of radius $r$	Dimensions $r \sqrt{2} \times \frac{r}{\sqrt{2}}$ ; area $= r^{2}$
Open box of max volume from square sheet by cutting corners	Cut size $= \frac{a}{6}$ from an $a \times a$ sheet

Worked Example — Window Optimisation (JEE type)

A window has the shape of a rectangle surmounted by a semicircle. Perimeter $= 10$ m. Find dimensions for maximum area.

Let width $= 2 r$ , rectangle height $= h$ .
Constraint: $2 h + 2 r + π r = 10$ → $h = \frac{10 - r (2 + π)}{2}$ .
Area: $A = 2 r h + \frac{π r^{2}}{2} = 10 r - r^{2} (2 + \frac{π}{2})$ .
$\frac{d A}{d r} = 10 - 2 r (2 + \frac{π}{2}) = 0$ → $r = \frac{10}{4 + π}$ .
Width $= 2 r = \frac{20}{4 + π}$ m; Height $= h = \frac{10}{4 + π}$ m.

4. Maxima and Minima Without Calculus — AM-GM Method JEE Advanced Only

For JEE Advanced, some optimisation problems are elegantly solved using the AM-GM inequality rather than calculus:

$\frac{a + b}{2} \geq \sqrt{a b}$ (equality when $a = b$ )

Example: Minimise $f (x) = x + \frac{1}{x}$ for $x > 0$ .

By AM-GM: $x + \frac{1}{x} \geq 2 \sqrt{x \cdot \frac{1}{x}} = 2$ , with equality when $x = \frac{1}{x}$ , i.e., $x = 1$ . Minimum value $= 2$ at $x = 1$ .

Standard Results Using AM-GM

$x + \frac{1}{x} \geq 2$ for $x > 0$ ; minimum at $x = 1$ .
$x^{2} + \frac{1}{x^{2}} \geq 2$ for $x \neq 0$ ; minimum at $x = \pm 1$ .
$a x + \frac{b}{x} \geq 2 \sqrt{a b}$ for $x > 0$ , $a, b > 0$ ; minimum at $x = \sqrt{b / a}$ .

Maxima/Minima — Complete Decision Flowchart

Situation	Method	Conclusion
$f^{'} (c) = 0$ , $f^{″} (c) \neq 0$	Second derivative test	$f^{″} (c) < 0$ : max; $f^{″} (c) > 0$ : min
$f^{'} (c) = 0$ , $f^{″} (c) = 0$	First derivative test (sign chart)	Sign change + to –: max; – to +: min; no change: inflection
Closed interval $[a, b]$	Closed interval method	Evaluate at all critical points AND endpoints
Optimisation with constraint	Single variable reduction + calculus or AM-GM	Verify nature + check domain boundaries

JEE Power Tips — Avoid These Common Mistakes

Every local extremum is a critical point, but not every critical point is an extremum. Always verify using the first or second derivative test after finding critical points. Never assume $f^{'} (c) = 0 \Rightarrow$ extremum.
$f^{″} (c) = 0$ does NOT mean a point of inflection. $f (x) = x^{4}$ has $f^{″} (0) = 0$ but $x = 0$ is a minimum. The sign of $f^{″}$ must change for it to be an inflection point.
For absolute extrema on a closed interval, always check endpoints. The absolute max/min could be at an endpoint, not at any interior critical point. Beginners forget to include endpoint values in the comparison.
In optimisation, always state the domain explicitly. Length must be positive, angle in $(0, π / 2)$ , etc. Missing this leads to extraneous solutions or incorrect identification of max vs min.
When the second derivative test fails ( $f^{″} (c) = 0$ ), switch to the first derivative test. Many students give up when $f^{″} (c) = 0$ — the first derivative test always works.
AM-GM is often faster than calculus for simple optimisation. For $a x + b / x$ type problems, directly apply AM-GM — it gives both the extreme value and the point in one line.

Practice Questions (JEE / Board Level)

Q1: The local maximum value of $f (x) = \sin x + \cos x$ on $[0, 2 π]$ is:

A) 1
B) $\sqrt{2}$
C) 2
D) $\frac{\sqrt{3}}{2}$

Answer: B) $\sqrt{2}$ .

Explanation:

First, find the critical points by setting the first derivative to zero:
$f^{'} (x) = \cos x - \sin x = 0$
$⟹ \tan x = 1$

Within the interval $[0, 2 π]$ , this occurs at $x = π / 4$ and $x = 5 π / 4$ .

Apply the second derivative test:
$f^{″} (x) = - \sin x - \cos x$

Evaluate at $x = π / 4$ :
$f^{″} (π / 4) = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \sqrt{2}$

Since $f^{″} (π / 4) < 0$ , there is a local maximum at $x = π / 4$ .

Calculate the maximum value:
$f (π / 4) = \sin (π / 4) + \cos (π / 4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$ .

Q2 (JEE Advanced type): A window has the shape of a rectangle surmounted by a semicircle. If the perimeter is 10 m, the width of the rectangle for maximum area is:

A) $\frac{10}{π + 4}$ m
B) $\frac{20}{π + 4}$ m
C) $\frac{10}{π + 2}$ m
D) $\frac{5}{π + 4}$ m

Answer: B) $\frac{20}{π + 4}$ m.

Explanation:

Let the width of the rectangle be $2 r$ (making the radius of the semicircle $r$ ) and the height of the rectangle be $h$ .

The perimeter consists of the semicircular arc, the bottom base, and the two vertical sides:
$P = 2 h + 2 r + π r = 10$
$⟹ 2 h = 10 - r (2 + π) ⟹ h = 5 - r (1 + \frac{π}{2})$

Write the total Area ( $A$ ) as the sum of the rectangle and semicircle areas:
$A = 2 r h + \frac{1}{2} π r^{2}$

Substitute $h$ into the area equation:
$A (r) = 2 r [5 - r (1 + \frac{π}{2})] + \frac{1}{2} π r^{2}$
$A (r) = 10 r - 2 r^{2} - π r^{2} + \frac{1}{2} π r^{2}$
$A (r) = 10 r - r^{2} (2 + \frac{π}{2})$

Differentiate with respect to $r$ and set to zero to maximize the area:
$A^{'} (r) = 10 - 2 r (2 + \frac{π}{2}) = 0$
$10 - r (4 + π) = 0 ⟹ r = \frac{10}{4 + π}$

The width of the rectangle is $2 r$ :
Width $= 2 (\frac{10}{4 + π}) = \frac{20}{π + 4}$ m.

Q3: Find the absolute maximum and minimum values of $f (x) = 2 x^{3} - 15 x^{2} + 36 x + 1$ on $[1, 5]$ .

A) Max 56, Min 24
B) Max 56, Min 1
C) Max 24, Min 1
D) Max 29, Min 24

Answer: A) Max 56, Min 24.

Explanation:

First, find the critical points by setting the derivative to zero:
$f^{'} (x) = 6 x^{2} - 30 x + 36$
$f^{'} (x) = 6 (x^{2} - 5 x + 6) = 6 (x - 2) (x - 3)$

The critical points are $x = 2$ and $x = 3$ . Both lie within the interval $[1, 5]$ .

Evaluate the function at the critical points and the endpoints:
$f (1) = 2 (1)^{3} - 15 (1)^{2} + 36 (1) + 1 = 24$
$f (2) = 2 (8) - 15 (4) + 36 (2) + 1 = 16 - 60 + 72 + 1 = 29$
$f (3) = 2 (27) - 15 (9) + 36 (3) + 1 = 54 - 135 + 108 + 1 = 28$
$f (5) = 2 (125) - 15 (25) + 36 (5) + 1 = 250 - 375 + 180 + 1 = 56$

Comparing these values, the absolute maximum is 56 (at $x = 5$ ) and the absolute minimum is 24 (at $x = 1$ ).

Q4: Using AM-GM, the minimum value of $\frac{4}{x} + x$ for $x > 0$ is:

A) 2
B) 4
C) 8
D) 1

Answer: B) 4.

Explanation:

According to the Arithmetic Mean-Geometric Mean (AM-GM) inequality, for any positive real numbers $a$ and $b$ , $\frac{a + b}{2} \geq \sqrt{a b}$ .

Let $a = \frac{4}{x}$ and $b = x$ . Applying AM-GM:
$\frac{\frac{4}{x} + x}{2} \geq \sqrt{\frac{4}{x} \cdot x}$

Simplify the geometric mean:
$\frac{\frac{4}{x} + x}{2} \geq \sqrt{4}$
$\frac{4}{x} + x \geq 2 (2) = 4$

The equality holds true when $\frac{4}{x} = x ⟹ x^{2} = 4 ⟹ x = 2$ . Therefore, the minimum value is 4.

Q5: At what point does $f (x) = x^{3} - 3 x$ have a local minimum?

A) $x = - 1$
B) $x = 0$
C) $x = 1$
D) $x = 3$

Answer: C) $x = 1$ .

Explanation:

First, find the critical points by taking the first derivative and setting it to zero:
$f^{'} (x) = 3 x^{2} - 3$
$3 (x^{2} - 1) = 3 (x - 1) (x + 1) = 0$

The critical points are $x = 1$ and $x = - 1$ .

Next, use the second derivative test:
$f^{″} (x) = 6 x$

Evaluate $f^{″} (x)$ at the critical points:
$f^{″} (1) = 6 (1) = 6 > 0$ (Concave up $⟹$ Local minimum)
$f^{″} (- 1) = 6 (- 1) = - 6 < 0$ (Concave down $⟹$ Local maximum)

Therefore, the local minimum occurs at $x = 1$ .

Q6: A closed cylindrical can of volume $128 π {cm}^{3}$ is to be made. Find the radius that minimises the total surface area.

A) $r = 2$ cm
B) $r = 4$ cm
C) $r = 8$ cm
D) $r = 16$ cm

Answer: B) $r = 4$ cm.

Explanation:

The volume of the cylinder is given as $V = π r^{2} h = 128 π$ .
Solving for height ( $h$ ):
$h = \frac{128}{r^{2}}$

The total surface area of a closed cylinder is $S = 2 π r^{2} + 2 π r h$ . Substitute $h$ into this equation:
$S (r) = 2 π r^{2} + 2 π r (\frac{128}{r^{2}})$
$S (r) = 2 π r^{2} + \frac{256 π}{r}$

To minimize the surface area, take the derivative with respect to $r$ and set it to zero:
$\frac{d S}{d r} = 4 π r - \frac{256 π}{r^{2}} = 0$
$4 π r = \frac{256 π}{r^{2}}$
$4 r^{3} = 256 ⟹ r^{3} = 64 ⟹ r = 4 cm$

Confirm it is a minimum using the second derivative:
$\frac{d^{2} S}{d r^{2}} = 4 π + \frac{512 π}{r^{3}}$
At $r = 4$ , $\frac{d^{2} S}{d r^{2}} > 0$ , confirming it is an absolute minimum.

Q7 (JEE Advanced type): Let $f (x) = x^{3} + b x^{2} + c x + d$ where $0 < b^{2} < c$ . Then $f (x)$ :

A) Has a local maximum in $(- \infty, 0)$ and a local minimum in $(0, \infty)$
B) Is monotonically increasing in $(- \infty, \infty)$
C) Has both a local maximum and a local minimum in $(- \infty, \infty)$
D) Is monotonically decreasing in $(- \infty, \infty)$

Answer: B) Is monotonically increasing in $(- \infty, \infty)$ .

Explanation:

Find the derivative to determine the monotonic behavior of the function:
$f^{'} (x) = 3 x^{2} + 2 b x + c$

This is a quadratic equation. Check its discriminant ( $Δ$ ) to see if it has any real roots:
$Δ = (2 b)^{2} - 4 (3) (c) = 4 b^{2} - 12 c = 4 (b^{2} - 3 c)$

We are given that $0 < b^{2} < c$ . Because $b^{2}$ is less than $c$ , it is strictly less than $3 c$ . Therefore, $b^{2} - 3 c$ must be a negative number.
$⟹ Δ < 0$

Since the discriminant is negative, the quadratic equation $f^{'} (x) = 0$ has no real roots. Furthermore, the leading coefficient of $f^{'} (x)$ is $3$ (which is positive), meaning the parabola opens upwards and never crosses the x-axis.

Therefore, $f^{'} (x) > 0$ for all real numbers $x$ . Because the derivative is always positive, the function $f (x)$ is strictly (monotonically) increasing on $(- \infty, \infty)$ and possesses no local extrema.

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Maxima, Minima & Optimisation

1. Local (Relative) Maxima and Minima

First Derivative Test

Second Derivative Test

Higher Order Derivative Test (when $f^{″} (c) = 0$ )

2. Absolute (Global) Maxima and Minima on $[a, b]$

Worked Example

3. Optimisation — Practical Maxima/Minima Problems

Classic Optimisation Results to Memorise

Worked Example — Window Optimisation (JEE type)

4. Maxima and Minima Without Calculus — AM-GM Method JEE Advanced Only

Standard Results Using AM-GM

Maxima/Minima — Complete Decision Flowchart

Practice Questions (JEE / Board Level)

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Maxima, Minima & Optimisation

1. Local (Relative) Maxima and Minima

First Derivative Test

Second Derivative Test

Higher Order Derivative Test (when f″(c)=0)

2. Absolute (Global) Maxima and Minima on [a,b]

Worked Example

3. Optimisation — Practical Maxima/Minima Problems

Classic Optimisation Results to Memorise

Worked Example — Window Optimisation (JEE type)

4. Maxima and Minima Without Calculus — AM-GM Method JEE Advanced Only

Standard Results Using AM-GM

Maxima/Minima — Complete Decision Flowchart

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Higher Order Derivative Test (when $f^{″} (c) = 0$ )

2. Absolute (Global) Maxima and Minima on $[a, b]$