Maxima and Minima — Optimizing Real Situations

Maxima and minima problems are asking a single question: at what input does a function reach its highest or lowest value? That might sound simple, but the technique — using derivatives — is one of the most widely applied ideas in mathematics. Engineers minimise material costs, physicists minimise energy, economists maximise profit, doctors minimise drug dosage. Calculus handles all of these through the same method.

Key Terms

Local maximum: A point $x = a$ where $f(a) \geq f(x)$ for all $x$ in some neighbourhood around $a$ . The function is higher here than at nearby points.

Local minimum: A point $x = a$ where $f(a) \leq f(x)$ for all $x$ in some neighbourhood around $a$ . The function is lower here than at nearby points.

Global (absolute) maximum: The highest value the function attains over its entire domain.

Global (absolute) minimum: The lowest value over the entire domain.

Critical point: A point where $f'(x) = 0$ or $f'(x)$ does not exist. Every local max/min is a critical point, but not every critical point is a local max/min (could be an inflection point).

Inflection point: A point where the function changes concavity (from concave up to concave down or vice versa). $f''(x) = 0$ at an inflection point, but this alone doesn’t make it a max or min.

Methods: Finding Maxima and Minima

Method 1 — First Derivative Test

Step 1: Find $f'(x)$ and solve $f'(x) = 0$ to get critical points.

Step 2: For each critical point $x = c$ , check the sign of $f'(x)$ on either side:

Sign of $f'(x)$	Conclusion
$+ \to -$ (positive left, negative right)	Local maximum at $x = c$
$- \to +$ (negative left, positive right)	Local minimum at $x = c$
No sign change	Inflection point (neither max nor min)

The logic: if $f'$ changes from $+$ to $-$ , the function was increasing then decreasing — it peaked at $c$ . If $f'$ changes from $-$ to $+$ , it was decreasing then increasing — it bottomed out at $c$ .

Method 2 — Second Derivative Test

After finding a critical point $x = c$ where $f'(c) = 0$ :

If $f''(c) > 0$ : local minimum (curve is concave up, like a bowl)
If $f''(c) < 0$ : local maximum (curve is concave down, like a hill)
If $f''(c) = 0$ : inconclusive — use the first derivative test instead

The second derivative test is faster in computation but fails when $f''(c) = 0$ .

Solved Examples

Example 1 — Basic (CBSE Class 12)

Find the local maxima and minima of $f(x) = x^3 - 3x + 2$ .

Step 1: $f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1)$

Critical points: $x = 1$ and $x = -1$ .

Step 2: Second derivative test: $f''(x) = 6x$

At $x = -1$ : $f''(-1) = -6 < 0$ → local maximum

$f(-1) = (-1)^3 - 3(-1) + 2 = -1 + 3 + 2 = 4$

At $x = 1$ : $f''(1) = 6 > 0$ → local minimum

$f(1) = 1 - 3 + 2 = 0$

Answer: Local max at $(-1, 4)$ ; local min at $(1, 0)$ .

Example 2 — Application: Maximise area (CBSE Class 12 / JEE Main)

A farmer has 200 m of fencing to enclose a rectangular field. Find the dimensions that maximise the area.

Set up: Let length = $x$ , width = $y$ . Constraint: $2x + 2y = 200$ , so $y = 100 - x$ .

Area: $A = xy = x(100 - x) = 100x - x^2$

Differentiate: $\frac{dA}{dx} = 100 - 2x$

Critical point: $100 - 2x = 0 \implies x = 50$

Check: $\frac{d^2A}{dx^2} = -2 < 0$ → maximum ✓

Answer: $x = y = 50$ m. A square gives maximum area: $A = 50 \times 50 = 2500$ m².

This is a classic result: among all rectangles with fixed perimeter, the square has maximum area.

Example 3 — Minimum surface area of cylinder (JEE Main level)

Find the dimensions of a cylinder with volume $V = \pi r^2 h = 1000\pi$ cm³ that has minimum total surface area.

Surface area: $S = 2\pi r^2 + 2\pi r h$

From the volume constraint: $h = 1000/r^2$

Substituting: $S = 2\pi r^2 + 2\pi r \cdot \frac{1000}{r^2} = 2\pi r^2 + \frac{2000\pi}{r}$

\frac{dS}{dr} = 4\pi r - \frac{2000\pi}{r^2}

Setting $= 0$ : $4\pi r = 2000\pi/r^2 \implies r^3 = 500 \implies r = \sqrt[3]{500} \approx 7.94$ cm

$h = 1000/r^2 = 1000/500^{2/3} \approx 15.87$ cm

Notice: $h = 2r$ (height equals diameter) — the optimal cylinder has height equal to diameter. This is a classic result examiners test.

Second derivative: $\frac{d^2S}{dr^2} = 4\pi + 4000\pi/r^3 > 0$ → minimum ✓

Example 4 — Hard (JEE Main level)

A window is in the shape of a rectangle surmounted by a semicircle. If the perimeter is 10 m, find the dimensions for maximum area.

Let the rectangle have width $2r$ (so the semicircle has radius $r$ ) and height $h$ .

Perimeter: $2r + 2h + \pi r = 10 \implies h = \frac{10 - 2r - \pi r}{2} = 5 - r - \frac{\pi r}{2}$

Area = rectangle + semicircle:

A = 2r \cdot h + \frac{\pi r^2}{2} = 2r\left(5 - r - \frac{\pi r}{2}\right) + \frac{\pi r^2}{2}

A = 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 = 10 - r(4 + \pi)

Setting $= 0$ : $r = \frac{10}{4 + \pi} \approx \frac{10}{7.14} \approx 1.40$ m

$h = 5 - r(1 + \pi/2) = 5 - 1.40(1 + 1.57) = 5 - 3.60 = 1.40$ m

So $h = r$ — the height of the rectangle equals the radius of the semicircle.

Exam-Specific Tips

CBSE Class 12: Application problems (area, volume, perimeter constraints) are standard 5–6 mark questions. The examiner requires: (1) setting up the function, (2) differentiating, (3) finding critical point, (4) proving it’s max/min using second derivative. All four steps must be shown. Verify with $d^2y/dx^2 > 0$ or $< 0$ — don’t just state the answer.

JEE Main: Optimisation problems appear as MCQs or numerical questions (1–4 marks). Speed matters. Master the setup: identify what to maximise/minimise, express it as a function of one variable using the constraint, differentiate, solve. Second derivative verification is rarely asked for in MCQs — just check the sign intuitively.

Common Mistakes to Avoid

Mistake 1 — Not checking the boundary. For a function on a closed interval $[a, b]$ , the global max/min might be at an endpoint, not a critical point. Always evaluate $f(a)$ and $f(b)$ in addition to the critical points, then compare all values.

Mistake 2 — Forgetting the domain. A critical point outside the physically meaningful domain is invalid. If $r$ is a radius, it must be positive. If $x$ is a length, it must satisfy $0 < x <$ maximum possible value.

Mistake 3 — Assuming $f''(c) = 0$ means inflection point. $f''(c) = 0$ means the second derivative test is inconclusive — it could still be a max, min, or inflection point. Use the first derivative test to determine the nature of the critical point.

Mistake 4 — Setting derivative of constraint equal to zero. The constraint is a fixed equation (like perimeter = 200). Don’t differentiate it to find critical points — it’s used to eliminate one variable. Only differentiate the objective function (area, volume, cost) after substitution.

Mistake 5 — Not verifying with the second derivative. In CBSE, showing that $d^2y/dx^2$ has the correct sign is required for full marks. Many students find the critical point and write the answer without proving it’s actually a maximum or minimum. This costs marks.

Practice Questions

Q1. Find the dimensions of a rectangle with perimeter 28 cm that has maximum area.

Let length = $x$ , width = $14 - x$ . Area $= x(14-x) = 14x - x^2$ . $dA/dx = 14 - 2x = 0 \implies x = 7$ . $d^2A/dx^2 = -2 < 0$ (maximum). Rectangle is a square with side 7 cm. Area = 49 cm².

Q2. Find the local maxima and minima of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$ .

$f'(x) = \cos x - \sin x = 0 \implies \tan x = 1 \implies x = \pi/4$ or $5\pi/4$ . $f''(x) = -\sin x - \cos x$ . At $x = \pi/4$ : $f'' = -\sin(\pi/4) - \cos(\pi/4) = -\sqrt{2} < 0$ → local max, $f(\pi/4) = \sqrt{2}$ . At $x = 5\pi/4$ : $f'' = \sqrt{2} > 0$ → local min, $f(5\pi/4) = -\sqrt{2}$ .

Q3. A box with a square base and no top is to be made from a square piece of cardboard by cutting equal squares from each corner. If the cardboard is 12 cm × 12 cm, find the size of the cut that maximises the volume.

Let cut size = $x$ cm. Base side = $12 - 2x$ , height = $x$ . Volume $V = (12-2x)^2 \cdot x$ . Expand: $V = x(144 - 48x + 4x^2) = 144x - 48x^2 + 4x^3$ . $dV/dx = 144 - 96x + 12x^2 = 12(12 - 8x + x^2) = 12(x-2)(x-6) = 0$ . Critical points: $x = 2$ or $x = 6$ (reject $x = 6$ as it gives zero base). At $x = 2$ : $d^2V/dx^2 = -96 + 24x = -96 + 48 = -48 < 0$ → maximum. Cut = 2 cm. Max volume = $(12-4)^2 \times 2 = 64 \times 2 = 128$ cm³.

Q4. Prove that $f(x) = x^3 - 3x^2 + 3x + 7$ has no local maxima or minima.

$f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x-1)^2 \geq 0$ for all $x$ . $f'(x) = 0$ only at $x = 1$ . Since $f'$ doesn’t change sign (it’s always $\geq 0$ ), $x = 1$ is an inflection point, not a max or min. The function is monotonically non-decreasing — no local extrema.

FAQs

Is every critical point a maximum or minimum? No. A point where $f'(x) = 0$ is only a candidate. It could be a local max, local min, or an inflection point (like $x = 0$ for $f(x) = x^3$ , where $f'(0) = 0$ but neither max nor min).

What if the second derivative is zero at a critical point? The second derivative test is inconclusive. Switch to the first derivative test: check the sign of $f'$ just to the left and right of the critical point.

Can a function have a global max without a local max? On a closed interval, the global max must occur either at a local max or at an endpoint. On an open interval or infinite domain, a function might not attain its global max at all (e.g., $f(x) = x$ on $(0, 1)$ has no maximum).

Why does the derivative equal zero at a max/min? At a smooth maximum, the tangent to the curve is horizontal — the function is momentarily flat before turning downward. A horizontal tangent has slope zero, so $f'(x) = 0$ . This is only true for smooth functions; a corner (non-differentiable point) can also be a max/min.

When should I use the first vs second derivative test? Use the second derivative test when it’s easy to compute $f''(c)$ and you’re sure it won’t be zero. Use the first derivative test when $f''(c) = 0$ , or when checking sign changes is faster than computing $f''$ .