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Name: Gisselle Washington Optimization problems Date: Oct. 23, z01q Period: 1 Prove Show that of all the rectangles with a given area, the one with the smallest perimeter is a square. hMimimun 1et A be a given 100f area 100 = Lw Suppose a cylindrical can is being made to hold 1 liter of oil. Design the can so that the minimum amount of metal can be used. The area of the metal is given by ar+2rh Note that 1L 1000cc so ar h = 1000, h 1000 1 2

Name: Gisselle Washington Optimization problems Date: Oct. 23, z01q Period: 1 Prove Show that of all the rectangles with a given area, the one with the smallest perimeter is a square. hMimimun 1et A be a given 100f area 100 = Lw Suppose a cylindrical can is being made to hold 1 liter of oil. Design the can so that the minimum amount of metal can be used. The area of the metal is given by ar+2rh Note that 1L 1000cc so ar h = 1000, h 1000 1 2

Calculus: Early Transcendentals
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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Name: Gisselle Washington Optimization problems Date: Oct. 23, z01q Period: 1 Prove Show that of all the rectangles with a given area, the one with the smallest perimeter is a square. hMimimun 1et A be a given 100f area 100 = Lw Suppose a cylindrical can is being made to hold 1 liter of oil. Design the can so that the minimum amount of metal can be used. The area of the metal is given by ar+2rh Note that 1L 1000cc so ar h = 1000, h 1000 1 2
Transcribed Image Text:Name: Gisselle Washington Optimization problems Date: Oct. 23, z01q Period: 1 Prove Show that of all the rectangles with a given area, the one with the smallest perimeter is a square. hMimimun 1et A be a given 100f area 100 = Lw Suppose a cylindrical can is being made to hold 1 liter of oil. Design the can so that the minimum amount of metal can be used. The area of the metal is given by ar+2rh Note that 1L 1000cc so ar h = 1000, h 1000 1 2
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