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Optimization of ellipse surrounding a circle

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Consider the ellipse ##\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1## that encloses the circle ##x^{2}+y^{2}=2x##. Find the values of a and b that minimize the area of the ellipse.

    2. Relevant equations
    ##Area=ab\pi##

    3. The attempt at a solution
    I begin by completing the square of the circle equation to get:

    ##(x-1)^2+y^2=1##

    I note that this circle is centered at (1,0). I know that a>b for a minimal area where the ellipse will touch the circle at 2 points, and if that is so, then a=x of the circle.

    I know I need to find a quadratic equation for x in terms of a and b by eliminating y^2. Then derive the ellipse and circle equation implicitly, and set them equal to each other.

    After an attempt of using these directions i get:

    An ellipse equation of
    ##\frac{x^{2}}{a^2}+\frac{-(x-1)^2+1}{b^2}=1##
    and the circle equation: ##y^2=-(x-1)^2+1##

    I hesitate to derive them because I think i'm missing a concept, but if these equations were correct I would derive them and set them equal to each other to find a relation between a and b.

    I am not sure if I am on wrong track, but please don't hesitate to tell me if I am misunderstanding a step.
     
  2. jcsd
  3. Oct 4, 2015 #2

    Ray Vickson

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    You have a problem of the form
    [tex] \begin{array}{l}\min_{a,b} \pi a b\\
    \text{subject to}\\
    \displaystyle \frac{x^2}{a^2} + \frac{1-(x-1)^2}{b^2} \leq 1 \; \forall x \in [0,2]
    \end{array} [/tex]
    This arises because any point ##(x,y)## on the circle must be inside or on the boundary of the elliptical region ##x^2/a^2 + y^2/b^2 \leq 1##.

    As written, you have infinitely many constraints, one for each ##x## in the range ##[0,2]##. In other words, you have infinitely many constraints of the form ##f(x,a,b) \leq 1## for all ##x##. Can you think of first solving an auxiliary optimization problem to turn this infinite number of constraints into a single constraint on ##a,b##? (Hint: it can be done.)

    Then you will have a simpler optimization problem of the form
    [tex] \begin{array}{l}\min_{a,b} \pi a b \\
    \text{subject to} \\
    g(a,b) = 0 \end{array} [/tex]
    for some relatively nice function ##g(a,b)##. At that point you can apply various methods, such as Lagrange multipliers, to finish the problem.
     
  4. Oct 5, 2015 #3
    How exactly would I solve an auxiliary optimization to turn into a single restraint?
     
  5. Oct 5, 2015 #4
    Also how do we know that the circle and ellipse are in the range [0,2] for x values?
     
  6. Oct 5, 2015 #5
    I think he's trying to say: If the LHS<=1 for all x in [0,2] , then in particular we have Max(over x) of the LHS <=1....
     
  7. Oct 5, 2015 #6
    I don't think I'm following that relationship either. How am I supposed to know the relationship between the ellipse and the circle.
     
  8. Oct 5, 2015 #7
    I know this problem is emulative of https://www.physicsforums.com/threa...area-of-an-ellipse-enclosing-a-circle.270437/ this one however I am just getting into multivariable differentiation so this is very confusing to me.
     
  9. Oct 5, 2015 #8
    How about since
    ##\frac{x^2}{a^2}+\frac{y^2}{b^2}=1##
    ##(x-1)^2+y^2=1##

    Can I set each side equal to each other or should I solve for y^2 of the circle equation to plug into the ellipse equation.
     
  10. Oct 5, 2015 #9

    Ray Vickson

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    Draw a circle of radius 1, centered at (1,0). What are the x-values you can have on the circle?

    I never, anywhere, said that the ellipse can extend only over 0 <= x <= 2; in fact, it can extend out to x = 10 million if you want it to. All I said was that the part of the ellipse lying between x = 0 and x = 2 must be outside the circle, and I put that mathematically in terms of an inequality constraint.

    As for telling you what to do next: I am not allowed to do that, by PF rules.
     
    Last edited: Oct 5, 2015
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