College Algebra / Alg 337 · Procedure · 60–90 seconds
Optimizing with a Quadratic
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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.
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.
Answer the question asked; the vertex hands over w, and the problem may want the area.
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