Applications of derivatives
The idea
Once you can differentiate, the derivative becomes an instrument: it locates maxima and minima, classifies the shape of graphs, drives related-rates problems, and powers approximations like linearization and Newton's method. The optimization workflow matters far beyond mathematics — minimize cost, maximize yield, find the most efficient design — and it all rests on one fact you can now prove: at an interior extreme point of a differentiable function, f'(x) = 0.
The mental model for optimization is a pipeline: model the quantity to optimize as a function of one variable (using the constraint to eliminate others), differentiate, solve f'(x) = 0 for critical points, then certify each candidate with the second-derivative test or a sign chart, and finally answer the actual question asked, with units. The most common error is stopping at f'(x) = 0. A vanishing derivative only flags a candidate — it might be a maximum, a minimum, or neither, and on a closed interval the endpoints compete too.
Worked example
A closed cylindrical can must hold 500 cm³. Find the radius r and height h that minimize the total surface area (top, bottom, and side).
- Write down the two ingredients: the constraint πr²h = 500 and the objective A = 2πr² + 2πrh (two disks plus the unrolled side).
- Use the constraint to eliminate h: h = 500/(πr²), so A(r) = 2πr² + 2πr × 500/(πr²) = 2πr² + 1000/r, a function of r alone on r > 0.
- Differentiate and set to zero: A'(r) = 4πr − 1000/r² = 0 gives r³ = 1000/(4π) = 250/π ≈ 79.58, so r ≈ 4.30 cm.
- Certify it is a minimum: A''(r) = 4π + 2000/r³ is positive for every r > 0, so the critical point is a genuine minimum — and the only one, since A blows up as r → 0 and as r → ∞.
- Recover the height: h = 500/(πr²) ≈ 500/(π × 18.50) ≈ 8.60 cm. Notice h ≈ 2r — the optimal can is exactly as tall as it is wide, a tidy structural check on the algebra.
Answer. The minimal-area can has r ≈ 4.30 cm and h ≈ 8.60 cm (height equal to the diameter), using about 349 cm² of material.
Check your understanding
- Why does f'(x) = 0 identify only candidates for extremes rather than guaranteeing one?
- How does the constraint equation earn its keep in an optimization problem with two natural variables?
- What changes in the workflow when the domain is a closed interval like 1 ≤ r ≤ 3 instead of r > 0?
- Where else have you seen the optimal shape of a design emerge from setting a derivative to zero, and why is that pattern so common?
Build the foundations first
Applications of derivatives builds on these concepts. If any feel shaky, start there.