Exercise · ZW-OPTIMIZATION-PROBLEMS-006 · ★☆☆ Basic
A rectangle has perimeter 48 m. What is the maximum possible area, in m²?
Choose one option.
Show answer
- D. \(144\)
Explanation
With width x, the length is \(24-x\) and the area is \(A(x)=x(24-x)\). \(A'(x)=24-2x=0\) gives x = 12, and the maximum area is \(A(12)=144\) (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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- ZW-OPTIMIZATION-PROBLEMS-008 What is the minimum value of f(x)=x^2-6x+13?
- ZW-OPTIMIZATION-PROBLEMS-009 What is a critical point of a function f?
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