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Rectangle inscribed in generic ellipse

  1. Oct 25, 2008 #1
    1. The problem statement, all variables and given/known data
    Largest possible area of a rectangle inscribed in the ellipse (x2/a2)+(y2/b2)=1

    2. Relevant equations
    Area of the rectangle = length*height


    3. The attempt at a solution
    I have it set up so that the four corners of the rectangle are at (x,y) (-x,y) (-x,-y) (x,-y) and that area therefore is A=(2x)(2y).
    In order to find the max area, I know I need to differentiate the equation, so I need to eliminate either x or y.
    Using the ellipse equation, I found x to be equal to sqrt(a2-(y2a2)/b2).
    Substituting that into the area equation I get:
    A=2(sqrt(a2-(y2a2)/b2))*(2y).

    And differentiating that has been a nightmare and I haven't gotten it right yet. I know a and b are constants, and become one in the derivative.

    But is there any easier way of solving this problem?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 25, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Looks to me like "Laplace multipliers" would make this much simpler. You have an ellipse of the form b2x2+ a2y2= a2b2. I notice you are assuming that the maximal area rectangle has its sides parallel to the axes of the ellipse. I'm not sure that follows easily but I can't see any more general way of doing this. You want to maximize A(xy)= xy with the constraint F(x,y)= b2x2+ a2y2= a2.
    [itex]\nabla A= y\vec{i}+ x\vec{j}[/itex] and [itex]\nabla F= 2b^2x\vec{i}+ 2a^2y\vec{j}[/itex].

    One must be a multiple of the other.
     
  4. Oct 25, 2008 #3
    I'm just in calculus BC and am not familiar with 'laplace multipliers' how does that work?
     
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