Calculus 1C — Applications of the Derivative

Optimization word problems

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

ELI5: "what's the best?" problems — biggest area, cheapest box, shortest time. The recipe: write the quantity to optimize, use a constraint to collapse it down to one variable, then set the derivative to zero.

A farmer has 100 ft of fence for a rectangular pen against a barn (the barn is the 4th side, no fence needed). If the width is w, the side parallel to the barn is 100 − 2w, so area A(w) = w·(100 − 2w). Then A′(w) = 100 − 4w = 0 ⇒ w = 25, giving 25 × 50 = 1250 ft².

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

pen area as a function of width

w_{opt} = 25

where A′ = 100 − 4w = 0

✓ pass

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

beats its neighbours — a genuine maximum

A_pen(x)

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

Real-world hook: optimization is the daily bread of engineering and business — least material for a can that holds a fixed volume, maximum strength for a fixed weight, minimum cost for a fixed output.

Try it yourself: with 100 ft of fence as a full four-sided rectangle (perimeter 100, so length = 50 − w), area is A(w) = w·(50 − w). Find the width that maximizes the area.

A_{rect}\left(w\right) = w \cdot \left(50 - w\right)

w_{best} = 25

✏️ Your turn: replace 5 with the width that maximizes w·(50 − w). The check only confirms it beats its neighbours — you have to find the peak.

✓ pass

A_{rect}(w_{best}) >= A_{rect}(w_{best} - 2) and A_{rect}(w_{best}) >= A_{rect}(w_{best} + 2)

green when your width really is the area-maximizing one