Optimizing Rectangle Dimensions for Area 1000m^2: A Simple Solution

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Homework Help Overview

The discussion revolves around an optimization problem involving a rectangle with a fixed area of 1000 m². Participants are tasked with finding the dimensions that minimize the perimeter.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between perimeter and area, with attempts to differentiate the perimeter equation. Some suggest plotting the perimeter to find minimum values, while others express uncertainty in their calculations and reasoning.

Discussion Status

The conversation includes various attempts to manipulate equations and differentiate to find critical points. There is a mix of confidence and uncertainty among participants, with some providing encouragement and others questioning their own understanding.

Contextual Notes

Participants express confusion and a need for clarification on differentiation concepts, indicating a potential gap in foundational knowledge that may affect their problem-solving approach.

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[SOLVED] simple optimization problem

Homework Statement



find the dimensions of a rectangle with area 1000 m^2 whose perimeter is as small as possible

Homework Equations



perimeter = 2x + 2y

area = xy

1000 = xy

y= 1000/x

perimeter = 2x + 2(1000/x)



The Attempt at a Solution



am stuck

any help from anybody?
 
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perimeter = 2x + 2(1000/x)
now differentiate!

should learn concepts from the book (if you don't know why you differentiate)

Simple approach:

plot your perimeter and pick and minimum value
this is realistic problem, so going for really large x and -ve x would be nonsense
 
2-(2000/x^2)=0

i think
 
physicsed said:
2-(2000/x^2)=0

i think
You think? ... have confidence!
 
2/1000=x^-2
 
no, try 2=2000/x^2
 
x=square root of .001?
 
x=\sqrt{1000}=\sqrt{10\cdot10^2}=10\sqrt{10}
 
x=\sqrt{1000}=\sqrt{10\cdot10^2}=10\sqrt{10}

how did u get that?
 
  • #10
2=\frac{2000}{x^2}\rightarrow x^2=\frac{2000}{2}\rightarrow x^2=1000
 
  • #11
thanks a lot. am messed up today!
 

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