Exercise · ZW-OPTIMIZATION-PROBLEMS-001 · ★☆☆ Basic
80 m of fencing is used to enclose a rectangular plot. What is the maximum possible area, in m²?
Choose one option.
Show answer
- A. \(400\)
Explanation
With width x, the length is \(40-x\) and the area \(A(x)=x(40-x)\). \(A'(x)=40-2x=0\) gives x = 20; the maximum area is \(A(20)=400\) (a square).
Review the concept: Optimization problems
An optimization problem seeks the largest or smallest value of a quantity subject to constraints, expressed as a function of one variable.
\[\text{maximize }A(x)\ \Rightarrow\ A'(x)=0\]
More practice problems: optimization problems
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