Drag anywhere: the object reshapes, and every size you try plots a point — your attempts trace out the quantity’s curve below. Hunt for the best one and your record is marked. Then press ✦ Solve it exactly: the derivative doesn’t guess — it lands the exact optimum, first try. See how close your hands got.
Every optimization problem — every single one — is the same six moves. Here they are on the fence problem, where the hardest move (killing the extra variable) shows itself most clearly. Reveal each step and read why before moving on.
The objective is the quantity you want biggest or smallest. The constraint is the fact that must stay true (so much fence, so much volume). The constraint is substitution fuel: solve it for one variable and feed it into the objective.
Lengths can’t be negative; a cut can’t exceed half the sheet. Write the domain first — it tells you which critical points are real and whether endpoints are candidates too.
f′(c)=0 makes c a candidate. Seal it: f″(c)<0 ⇒ max (or >0 ⇒ min) — and if c is the only critical point in the domain, it’s the absolute winner.
x = 25 is where the best happens — not the answer. If they asked for the maximum area, report A = 1250 m². Reread the question, give the right quantity, with units.
If your formula still has two variables, stop. Use the constraint first — you can only take \(\frac{d}{dx}\) of a one-variable function.
A critical point outside the domain is garbage (a 6-inch cut from a 12-inch sheet). And on a closed interval, the endpoints are candidates too.
“x = 25” earns setup points, not the answer point. Report the quantity they asked for — the area, the volume, the cost — with units.
Finding \(f'=0\) and declaring victory skips the test. Show the sign change, or \(f''\), or the candidates table — the AP reader needs the why.