College Algebra / Unit 6 Alg 337 worked example, 60 to 90 seconds
Optimizing with a Quadratic
The idea
Optimizing with a quadratic means writing the quantity as a quadratic function of one variable, locating the vertex, and reading the extreme value with its units.
In practice
Forty meters of fence enclose a rectangle against a long wall; the wall needs no fence. With width w on the two sides, the length is 40 − 2w, and the area is A(w) = w(40 − 2w) = 40w − 2w². The parabola opens down — a = −2 — so the vertex is a maximum: w = −40/(2 · −2) = 10, and A(10) = 400 − 200 = 200. The answer with units: width 10 meters, length 20, area 200 square meters. The plain-sense audit runs at the neighbors: A(9) = 198 and A(11) = 198, both under 200 — the peak is where the formula said.
Where people slip
Answer the question asked; the vertex hands over w, and the problem may want the area.
Your place on the trail
Every lesson rests on earlier ones and opens later ones. Tap any to jump.
Builds on
- Nothing. This is a starting point.
Unlocks
- Nothing yet depends on this.
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