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{Calculus 1} Optimization Problem

  1. Aug 1, 2015 #1
    1. The problem statement, all variables and given/known data
    A wire is divided into two parts. One part is shaped into a square, and the other part is shaped into a circle. Let r be the ratio of the circumference of the circle to the perimeter of the square when the sum of the areas of the square and circle is minimized. Find r.

    2. Relevant equations
    Derivative and area of circle and square

    3. The attempt at a solution

    So I figure I have a wire y since I don't know the length cut it into two pieces x for the square and y-x for the circle. Area of the square is (x/4)^2 While y-x=2pir, r=y-x/2pi So area of circle is (y-x/2pi)^2

    Now I want to minimize the total area by taking the derivative and setting it equal to zero. My problem is the y and this doesn't seem right. Any help is appreciated I still struggle with these word problems
     
  2. jcsd
  3. Aug 1, 2015 #2
    What is the objective function?

    What is the constraint function?

    Define those first.
     
  4. Aug 1, 2015 #3
    I messed up the area of the circle

    so I want to minimize



    area of square + area of circle = x^2/16 + (y-x)^2/pi

    Which would be my constraint

    Objective function

    In order to get x and y for r = circumference of circle / perimeter of square = (y - x) / x

    Do I have all the information I need now my problem is if I take the derivative of that and set it equal to zero I have two variables
     
    Last edited: Aug 1, 2015
  5. Aug 2, 2015 #4

    Ray Vickson

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    (1) You do not have two variables; a poor choice of notation has allowed you to mislead yourself. The wire is given, so its length is not variable. Of course, you do not happen to be told its exact length, but for purposes of optimization that is irrelevant: the length is an input parameter, not a variable. Or, to put in another way: if the length of the wire is also regarded as a variable, the best solution is to take length = 0, as that certainly minimizes the total of both areas.
    (2) In optimization, the objective function is the thing you are trying to maximize or minimize, while constraints are restrictions on the variables.
    (3) Your formula for the area of the circle is wrong.
     
  6. Aug 2, 2015 #5
    Yes thank you your first point was my biggest mistake I was making. Its funny how understanding the question correctly makes the process much easier.

    I fixed the area equation took the derivative set it equal to zero and finally got the solution of pi/4 as the correct answer
     
  7. Aug 2, 2015 #6

    Ray Vickson

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    Correct!
     
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