Calculus 1 — Limits & Derivatives

Using it: optimization

continues from lesson 3 — values defined earlier in the course stay live here

Maximums and minimums live where the slope is zero. A farmer has 100 m of fence for a rectangular pen against a barn (no fence needed on that side): area A(w) = w·(100 − 2w). The derivative 100 − 4w is zero at w = 25:

A_{pen}\left(w\right) = w \cdot \left(100 - 2 \cdot w\right)

area as a function of width

w_{best} = \frac{100}{4}

25

where the derivative vanishes

A_{max} = \mathrm{A_{pen}}\left(w_{best}\right)

1250

1250 m² — the best possible

✓ pass

A_{pen}(w_{best}) > A_{pen}(w_{best} - 1) and A_{pen}(w_{best}) > A_{pen}(w_{best} + 1)

beats its neighbors — it really is a peak

A_pen(x)

pen area vs width — the peak sits exactly at w = 25

Where next: Calculus 2 runs the machine in reverse — from slopes back to areas.