On this page
Most students can differentiate x³ - 3x. Far fewer can turn "a box is made by cutting squares from the corners of a sheet" into a function to differentiate. Applications of derivatives are tested on exactly that translation step. This post gives you a set-up routine for optimisation and for related rates, three worked problems with full numbers, and the slips that cost marks.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists, under Unit 7, "Applications of derivatives: Rate of change of quantities, monotonic-Increasing and decreasing functions, Maxima and minima of functions of one variable." The NCERT Class 12 chapter on application of derivatives covers rates of change, increasing and decreasing functions, the first and second derivative tests, and absolute maxima and minima on closed intervals. Check the syllabus for your exam year.
Why are these problems hard?
The calculus is usually the easy part. Tami Martin (2000) studied university students in an introductory calculus course solving related-rates problems set in geometric contexts. Overall performance was poor, and it was poorest on the steps linked to conceptual understanding. In practice, that is the set-up: turning the words into a diagram and an equation. So practise the set-up as a skill on its own.
How do you set up an optimisation problem?
Use the same five steps each time:
Draw and label the situation. Name the variable you will control.
Write the quantity to optimise (volume, area, cost) as a function of that one variable. Use the given constraint to remove any second variable.
State the allowed interval for the variable.
Differentiate, set to zero and solve for critical points.
Test each point with the first or second derivative test, and on a closed interval compare the endpoint values too.
Worked example 1: the open box
A square sheet of side 18 cm has equal squares of side x cut from each corner, and the sides are folded up to make an open box. Find x for maximum volume.
The base is (18 - 2x) by (18 - 2x) and the height is x, so V(x) = x(18 - 2x)².
The interval: 0 < x < 9, since the base must have positive length.
Expand: V = 4x³ - 72x² + 324x. Differentiate: V'(x) = 12x² - 144x + 324 = 12(x² - 12x + 27) = 12(x - 3)(x - 9).
Critical points: x = 3 or x = 9. Only x = 3 lies inside the interval.
V''(x) = 24x - 144. At x = 3, V'' = -72 < 0, so x = 3 gives a maximum.
Maximum volume: V(3) = 3 × 12² = 432 cm³.
Worked example 2: endpoints matter
Find the absolute maximum and minimum of f(x) = x³ - 3x on [0, 2].
f'(x) = 3x² - 3 = 0 gives x = ±1. Only x = 1 is in the interval.
Evaluate at the critical point and both endpoints: f(0) = 0, f(1) = 1 - 3 = -2, f(2) = 8 - 6 = 2.
Absolute maximum: 2, at the endpoint x = 2. Absolute minimum: -2, at x = 1.
A student who only checks f'(x) = 0 reports a local minimum and misses the maximum, which sits at an endpoint where the derivative is not zero.
How do you solve a related-rates problem?
Here the quantities change with time, and you want the rate of one from the rate of another.
Draw and label, using letters for quantities that change and numbers only for those that stay fixed.
Write an equation linking the quantities, true at every moment.
Differentiate both sides with respect to time, using the chain rule.
Only now substitute the values at the instant asked about.
Worked example 3: the sliding ladder
A 5 m ladder leans against a vertical wall. Its foot slides away from the wall at 0.5 m/s. How fast is the top sliding down when the foot is 3 m from the wall?
Let x be the distance of the foot from the wall and y the height of the top. The ladder length is fixed: x² + y² = 25.
Differentiate with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.
At the instant asked, x = 3, so y = √(25 - 9) = 4. Also dx/dt = 0.5 m/s.
Substitute: 3 × 0.5 + 4 × (dy/dt) = 0, so dy/dt = -1.5 / 4 = -0.375 m/s.
The top is sliding down at 0.375 m/s. The negative sign means y is decreasing.
If you had put x = 3 into the equation before differentiating, you would have treated a changing length as a constant, and the derivative would have been wrong. That single ordering slip is the most common related-rates error.
Common slips and how to avoid them
Keeping two variables in the function to optimise instead of using the constraint.
Ignoring the allowed interval, such as accepting x = 9 in the box problem.
Forgetting the endpoints on a closed interval.
Treating f'(c) = 0 as proof of an extremum.
Substituting values before differentiating in related rates.
Dropping the sign of a rate, or its units.
A practice routine
Set-up drill: for ten word problems, write only the function and its interval. Do not differentiate yet. Check the set-ups first.
Closed-interval drill: five functions where the answer sits at an endpoint.
Related rates: five problems with ladders, cones of water, expanding circles and spheres, written with the four-step method.
Mixed timed set of ten from NCERT exercises and past papers.
For the calculus behind these steps in physics questions, read our post on the calculus you need for JEE physics. If differentiability itself is unclear, start with limits and continuity. When a set-up will not come, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, write the function and interval for three optimisation word problems without differentiating, then check each against the solution.
