Exercise · ZW-OPTIMIZATION-PROBLEMS-002 · ★☆☆ Basic
120 m of fencing is used to enclose a rectangular plot. What is the maximum possible area, in m²?
Choose one option.
Show answer
- D. \(900\)
Explanation
With width x, the length is \(60-x\) and the area \(A(x)=x(60-x)\). \(A'(x)=60-2x=0\) gives x = 30; the maximum area is \(A(30)=900\) (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
- ZW-OPTIMIZATION-PROBLEMS-001 80 m of fencing is used to enclose a rectangular plot. What is the maximum possible area, in m²?
- ZW-OPTIMIZATION-PROBLEMS-003 Two positive numbers have product 36. What is the smallest possible value of their sum?
- ZW-OPTIMIZATION-PROBLEMS-004 Equal squares are cut from the corners of a 12 cm square sheet and the sheet is folded into an open…
- ZW-OPTIMIZATION-PROBLEMS-005 What is the maximum value of f(x)=-x^2+8x+5?
- ZW-OPTIMIZATION-PROBLEMS-006 A rectangle has perimeter 48 m. What is the maximum possible area, in m²?
- ZW-OPTIMIZATION-PROBLEMS-007 Two positive numbers add up to 20. What is the largest possible value of their product?
- 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?
← ZW-OPTIMIZATION-PROBLEMS-001 ZW-OPTIMIZATION-PROBLEMS-003 →