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The Circle and the Square - How do I get a max value from a min value problem?

  1. Sep 24, 2007 #1
    1. The problem statement, all variables and given/known data
    A piece of wire 50 cm long is to be cut into two pieces. (X and 50-x)
    One piece (X) will form a circle; the other (50-x) will be bent to form a square.
    Two students, Max and Min, have different goals. Min wants to minimize the sum of the areas of the circle and square. Max wants to maximize the sume of the areas of the circle and square.

    There must only be one (an) equation that will be minimized and maximized.

    Hint: You must consider the domain carefully in this situation [D(0,50) ?]
    2. Relevant equations
    Circumference:X = 2*pi*R
    Area of the Circle:A = pi*R^2
    Area of the Square: A = S^2
    Length of a side of a square: S = (50-X)/4

    3. The attempt at a solution

    First, I found the minimum of the problem:
    Circumference -
    X = 2*pi*R
    X/(2*pi) = R

    A = pi*R^2
    A = pi*(x/(2*pi))^2
    Length of a side: 4s = (50-x)
    s = (50-x)/4

    Area = S^2
    Area = ((50-x)/(4))^2

    Total Area (T.A) = A of Circle + A of Square
    T.A = pi*(x/(2*pi))^2 + ((50-X)/4)^2
    T.A = (4*x^2 + pi*(50-x)^2))/(16*pi)

    T.A is approximately 87.52 with x approximately being 22. The answers were obtained by using my graphing calculator and rounding to the hundredths.

    The domain I am guessing for this function of T.A is 0<x<50.

    But, I haven't got the slightest clue on how to get a maximum from this problem... Any help would be appreciated.
  2. jcsd
  3. Sep 25, 2007 #2


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    I would say the domain is 0<=x<=50, you should be allowed to put all of the wire into one shape if you want to achieve the max. Which one should you put all of it into (i.e. look at the endpoints of the domain as I think the hint is suggesting).
  4. Sep 25, 2007 #3


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    Don't forget that, because you are working with a *finite* domain for x, the critical point that you solved for only represents the value of x where d(T.A.)/dx = 0 in the interval. That is the local minimum for your total area function.

    Have you learned about *global* or absolute minima and maxima of a function on an interval? Check the value of T.A. at the endpoints of your interval and compare them with your critical value and with each other.
  5. Sep 25, 2007 #4


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    If f(x) is given on an interval [a, b], both max and min must occur:
    1) in the interior (a,b) where f '(x)= 0 or
    2) in the interior (a,b) where f '(x) does not exist (not relevant here) or
    3) at an endpoint.

    Find the points where f '(x)= 0, evaluate f at those points and a and b and pick out the highest and lowest.
  6. Sep 25, 2007 #5
    Unfortunately, we have not yet learned derivatives.... Which is what is 'making this problem' hard.

    I've decided to just call 49 my local maximum instead of doing 49.9999 repeating. Any other comments / suggestions would be welcome...

    Thanks for the help so far!
  7. Sep 25, 2007 #6


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    If you are considering only integer values of x that are not equal to zero or fifty, that's correct. Otherwise it's completely wrong. 49.1 is bigger. What's stopping you from saying the max is at x=50?
  8. Jun 30, 2009 #7
    Dynamicsolo, here is my two cents eventho I have not done max and min problems for years.
    Plot you equation of TA=f(x) in the interval of 0<x<50 and see how it curves. We know that is is a parabolic curve and the max value should be evident but not necessarily useful. If you can use excel, it can plot for you quickly.
  9. Jul 5, 2009 #8
    have you covered maxima & minima yet?
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