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CalculusCalculus 1College

Applications of Derivatives

Extrema, concavity, related rates and optimization.

Because the derivative measures slope, its sign tells you where a function rises (f′>0f' > 0) or falls (f′<0f' < 0). Points where f′=0f' = 0 or f′f' is undefined are critical points, the only candidates for local maxima and minima. The second derivative describes concavity and confirms which candidates are peaks and which are valleys.

These ideas solve optimisation problems, such as the largest area for a given fence length or the cheapest can for a given volume, and related-rates problems, where you differentiate an equation linking quantities with respect to time.

Key ideas

  • First derivative test: a sign change from ++ to −- marks a local maximum; −- to ++ a local minimum.
  • Second derivative test: f′′(c)<0f''(c) < 0 at a critical point means a maximum, f′′(c)>0f''(c) > 0 a minimum.
  • On a closed interval, compare the function at critical points and at both endpoints.
  • Inflection points occur where concavity changes.

Worked example

A farmer has 200 m of fence to enclose a rectangle against a river (no fence on the river side). What dimensions maximise the area?

  1. Let xx be each side perpendicular to the river; the parallel side is 200−2x200 - 2x.
  2. Area A(x)=x(200−2x)=200x−2x2A(x) = x(200 - 2x) = 200x - 2x^2.
  3. A′(x)=200−4x=0A'(x) = 200 - 4x = 0 gives x=50x = 50.
  4. A′′(x)=−4<0A''(x) = -4 < 0, so a maximum: 5050 m by 100100 m, area 50005000 m².

Answer: 5050 m by 100100 m

Common mistakes

  • Assuming every critical point is a maximum.
  • Ignoring endpoints on a closed interval.
  • Answering with xx only when the question asks for the dimensions or the maximum value.

Practice questions

Try each one, then open the answer. Numerical answers on this page are verified by an automated calculation check.

  1. 1.Find the xx-coordinate of the minimum of f(x)=x2−6x+2f(x) = x^2 - 6x + 2.

    Show answer

    33

  2. 2.Find the maximum value of f(x)=−2x2+8xf(x) = -2x^2 + 8x.

    Show answer

    88

  3. 3.Find the critical point of f(x)=ex−2xf(x) = e^x - 2x.

    Show answer

    x=ln⁡2≈0.693x = \ln 2 \approx 0.693

Keep practising

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