A piece of wire 110 meter long is cut into two. One piece of length z is bent into a circle while the other piece into a square. Find the minimum area enclosed by the two shapes. Student solution. One piece of length z is bent into a circle. Then in tems of z, the radius of the circle is given by The area for the circle piece is then given by Acircle =| The length of the second piece in terms of z is given by second piece = The side of the square is then given by The area for the square piece is then given by Asquare Denote by A(r) the combined area. Then it follows that A(x) = The derivative of A(x) is given by A'(x) = After solving the appropriate equation, it then follows that the minimum area is atained at Imin
A piece of wire 110 meter long is cut into two. One piece of length z is bent into a circle while the other piece into a square. Find the minimum area enclosed by the two shapes. Student solution. One piece of length z is bent into a circle. Then in tems of z, the radius of the circle is given by The area for the circle piece is then given by Acircle =| The length of the second piece in terms of z is given by second piece = The side of the square is then given by The area for the square piece is then given by Asquare Denote by A(r) the combined area. Then it follows that A(x) = The derivative of A(x) is given by A'(x) = After solving the appropriate equation, it then follows that the minimum area is atained at Imin
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:A piece of wire 110 meter long is cut into two. One piece of length z is bent into a circle while the other piece into a square. Find the minimum area enclosed by the two shapes.
Student solution. One piece of length z is bent into a circle. Then in terms of z, the radius of the circle is given by
The area for the circle piece is then given by
Acircle =
The length of the second piece in terms of z is given by
second piece =
%3D
The side of the square is then given by
The area for the square piece is then given by
Asquare
Denote by A(r) the combined area. Then it follows that
A(2) =
The derivative of A(x) is given by
A'(x) =
After solving the appropriate equation, it then follows that the minimum area is atained at
Imin
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