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calculus help needed with Optimization please :)?
A piece of wire x m long is cut into two pieces, the length of the first piece being 18 m. The first piece is bent into a circle, and the other is bent into a rectangle with length twice the width. Give an expression for the total area A enclosed in the two shapes in terms of x.
1 Answer
- AshLv 77 years ago
The first piece of 18m is bent to for circumference of a circle
circumference 18 = 2πr (r = radius of the circle)
r = 18/2π = 9/π
Area = π r^2 = π (9/π)^2 = 81/π
For 2nd piece the remaining length is 18-x
Since rectangle length is twice the width, we have width = (18-x)/ 6 and length as (18-x)/3
Area of rectangle = (18-X)/6 * (18-x)/3 = (18-x)^2 / 18
Total Area = 81/π + [(18-x)^2 /18]
