Skill goal: Model an optimization problem, find critical points, and verify a maximum.
1 · Observe
Two classic problems. Pick a tab and look at the setup before touching anything.
Fence problem: you have 40 m of fence to enclose a rectangular pen against a barn, so the barn covers one long side. If the width is x (0 < x < 20), the length is 40 − 2x and the area is A(x) = 40x − 2x².
Box problem: cut equal squares of side x from the corners of a 12 × 12 sheet and fold up an open box. The volume is V(x) = x(12 − 2x)², for 0 < x < 6.
2 · Manipulate
Drag the slider. The marker on the graph moves and the readouts update live.
3 · Predict
Question: which fence width x maximizes the pen’s area? Type your guess, then check.
4 · See
Live readouts at the current slider position:
5 · Explain
Fence: A′(x) = 40 − 4x. Setting A′(x) = 0 gives the critical point x = 10. By the first derivative test, A′ changes sign there: for x < 10, A′(x) > 0 (area increasing); for x > 10, A′(x) < 0 (area decreasing). A sign change from + to − means a local maximum — and on (0, 20) it is the global maximum: A(10) = 400 − 200 = 200 m².
Box: V(x) = x(12 − 2x)², so V′(x) = (12 − 2x)² + x·2(12 − 2x)(−2) = (12 − 2x)(12 − 6x). Critical points: x = 2 and x = 6. On (0, 6), x = 6 gives volume 0 (minimum), while x = 2 gives the maximum. Verification: V(2) = 2·(12 − 4)² = 2·64 = 128.
6 · Challenge
Three questions, auto-checked.
Score: 0 / 3