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Daily Math Minute

Unit 5: Analytical Applications of Differentiation

Optimization

Solving applied optimization problems using derivatives.

Advanced25 min lesson2 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

  • First & Second Derivative Tests

Minimizing Material for a Fixed Volume

An optimization problem starts by translating a constraint into a single-variable function, then applies a derivative test to its critical points.

Worked Example — Minimizing a Cylinder's Surface Area

A closed cylindrical can must hold V = 250π cm³. Minimize the material used. With radius r and height h: V = πr²h = 250π, so h = 250/r². Surface area (top, bottom, and side): S = 2πr² + 2πrh = 2πr² + 500π/r. dS/dr = 4πr − 500π/r²; setting this to 0 gives 4r³ = 500, so r³ = 125, r = 5. Then h = 250/25 = 10. Confirm a minimum: d²S/dr² = 4π + 1000π/r³, positive for every r > 0, so S is concave up everywhere and this single critical point is an absolute minimum. S(5) = 2π(25) + 500π/5 = 50π + 100π = 150π cm². Check the volume constraint: πr²h = π(25)(10) = 250π — matching.

Tip

When the second derivative has a fixed sign across the whole domain (as here, d²S/dr² > 0 for every r > 0), a single critical point is guaranteed to be an absolute extremum — no endpoint comparison is needed.

Common Mistakes

  • Optimizing r without first eliminating the second variable, h, using the given constraint.

    A constrained optimization needs a single-variable function before any derivative can be taken — h = 250/r² is exactly what eliminates h from the surface-area formula.

Key Takeaways

  • Optimization starts by using a given constraint to write the quantity to optimize as a function of one variable.
  • A domain-wide sign on the second derivative confirms an absolute extremum without checking endpoints separately.

Summary

This closes Unit 5. Unit 6 turns to the opposite question — recovering accumulated quantities from a rate of change, starting with approximating and then exactly defining area under a curve.