Maxima and Minima Applications for JEE Main 2026

Application of derivatives is the most practically demanding part of JEE Main calculus, requiring not just differentiation skill but the ability to set up optimisation problems from verbal or geometric descriptions. Maxima and minima questions appear in three to four questions per session and reward students who have a systematic routine for finding critical points, testing them, and extracting the answer that matches the geometric or physical context.

First Derivative Test

At a critical point x = c (where f'(c) = 0 or f'(c) is undefined): if f'(x) changes from positive to negative at c, then f(c) is a local maximum. If f'(x) changes from negative to positive, it is a local minimum. If f'(x) does not change sign, it is an inflection point (not an extremum). The first derivative test requires finding all critical points and testing the sign of f' on either side. For a polynomial, factor f'(x) and use a sign chart. For the derivatives foundation see our application of derivatives guide.

Second Derivative Test

At a critical point x = c where f'(c) = 0: if f''(c) > 0, then f(c) is a local minimum (concave up); if f''(c) < 0, it is a local maximum (concave down); if f''(c) = 0, the test is inconclusive (use the first derivative test). The second derivative test is faster when f'' is easy to compute and non-zero at the critical point. For f(x) = x³ − 3x + 2: f'(x) = 3x² − 3 = 0 at x = ±1. f''(x) = 6x: f''(1) = 6 > 0 (minimum at x = 1, value = 0); f''(−1) = −6 < 0 (maximum at x = −1, value = 4).

Absolute Extrema on a Closed Interval

On a closed interval [a, b], the absolute maximum and minimum occur at either the interior critical points or the endpoints. The procedure: (1) find all critical points in (a, b), (2) evaluate f at each critical point and at a and b, (3) identify the largest (absolute maximum) and smallest (absolute minimum) values. JEE applied problems often ask for the maximum area, minimum perimeter, or minimum cost — always set up the function, find the feasible domain, and apply this procedure.

Applied Optimisation: Standard Problem Types

The cylinder of maximum volume inscribed in a sphere: express V in terms of one variable using the sphere constraint, differentiate, set to zero. The rectangle of maximum area inscribed in a circle: set up using the half-angle parametrisation. The right triangle of maximum area with fixed hypotenuse: maximise ½ab subject to a² + b² = c². In all cases, the routine is identical — express the objective in one variable, differentiate, solve. Take a free mock test to practise setting up these geometric optimisation problems.

AM-GM as a Non-Calculus Shortcut

For sums of positive terms, the AM-GM inequality often finds extrema instantly: x + 1/x ≥ 2√(x · 1/x) = 2, with equality at x = 1. Similarly, a²x + b²/x is minimised at x = b/a with minimum value 2ab. Recognising an AM-GM structure bypasses all differentiation. This shortcut deserves priority over calculus when the expression is a sum of reciprocal-type terms — faster and less error-prone. After mastering both approaches, integrate with our mathematical inequalities guide for the AM-GM connection.