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Find the dimensions of the rectangle of greatest area

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the dimensions of the rectangle of greatest and least area
    that can be inscribed in the ellipse x^2/16 + y^2/9 = 1 with sides parallel
    to the coordinate axes.

    3. The attempt at a solution
    f(x,y) = (2x)(2y) = 4xy
    ∇f = <4y,4x>
    ∇g = <x/8,2y/9>

    ∇f = λ∇g

    4y = λx/8
    4x = λ2y/9

    I can isolate lambda because y=0,x=0 is not a part of the ellipse.

    λ = 32y/x
    λ = 18x/y

    32y/x = 18x/y
    x^2 = 16y^2/9

    Sub into ellipse

    2y^2/9= 1

    y = 3/√2
    x = 2√2

    I don't know what to do after this, I know I am supposed to check whether it is a maximum or not, but I don't know the method. What would the next step be?
     
  2. jcsd
  3. Nov 14, 2014 #2

    Zondrina

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    This problem requires a little bit more work than you would expect. You have these equations:

    ##4y = \frac{\lambda x}{8}##
    ##4x = \frac{\lambda 2y}{9}##
    ##\frac{x^2}{16} + \frac{y^2}{9} - 1 = 0##

    You're going to wind up finding four different points to test. Two will give you a max and two will give you a min.

    For the first two points why don't you try looking at the first two equations and solving for ##x## like so:

    ##x = \frac{32y}{\lambda}##
    ##x = \frac{\lambda y}{18}##

    This should give ##\lambda = 24##. Subbing back it's easy to see ##y = \frac{3}{4} x##. Using the third equation now should give you two ##x## points, which will yield two ##y## points when subbed in. These are two of the points you need.

    Now go back to the first two equations. Solve for ##y## and repeat the above procedure. This will give you two more points to test.
     
  4. Nov 14, 2014 #3
    Oh yea, I know what I did wrong, since its a closed loop, you can just check the max of the points you get. I was thinking it wasn't a closed loop and you had to do something with end behaviour.
     
  5. Nov 15, 2014 #4

    Ray Vickson

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    Can you be sure that the Lagrange derivative conditions hold at both the maximum and the minimum?
     
  6. Nov 15, 2014 #5

    Ray Vickson

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    He will not get four feasible points to test---only two. (There is something seriously wrong with the other two "points"). This is why I asked the OP the question in post #4.
     
    Last edited: Nov 15, 2014
  7. Nov 15, 2014 #6
    If I am not sure, that the lagrange derivative does that. What would I do differently?
     
  8. Nov 15, 2014 #7

    haruspex

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    How do you think area is defined for the purposes of this question? Can it be negative?
     
  9. Nov 15, 2014 #8

    Ray Vickson

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    If (x,y) is a corner of the rectangle, the area ##= 4 |x| |y|##. The function ##f(x,y) = 4xy## represents the area only in the first and third quadrants. Anyway, to get rid of false solutions and ambiguities, it is better to formulate the problem as
    [tex] \max / \min \: f(x,y) = 4 x y\\
    \text{subject to } \: g(x,y) \equiv \frac{x^2}{16} + \frac{y^2}{9} = 1\\
    \text{and} \; x,y \geq 0 [/tex]
    A solution at ##(x,y)## is accompanied by three other mirror-image points that also solve the problem.

    The sign-constraints on ##x,y## introduce some changes to the Lagrangian conditions: if
    [tex] L = f(x,y) -\lambda g(x,y) [/tex]
    we have:
    1. For a max problem, [tex] \partial L / \partial x \leq 0, \; \text{ and} \; \partial L / \partial x = 0 \; \text{if} \; x > 0\\
      \text{same for } \; y [/tex]
    2. For a min problem, [tex]
      \partial L / \partial x \geq 0, \; \text{ and} \; \partial L / \partial x = 0 \; \text{if} \; x > 0\\
      \text{same for } \; y
      [/tex]
    So, conditions are a bit different if x = 0 or y = 0.
     
  10. Nov 15, 2014 #9
    Time to look through my textbook now, everything you said is pretty foreign, my professor didn't go over that. Basically my professor introduced it as, thinking of the constraint and the original function as level curves, and we are trying to find the points where they touch, so if their tangents are parallel so are the gradients, then
    ∇f = λ∇g
    I think he showed us that you get 4 points, and to basically plug each one of them into the original to figure out the max/min.
     
  11. Nov 15, 2014 #10

    Ray Vickson

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    You do get four points in the Lagrian problem: one at x>0, y >0 and the mirror image of that. You get two other solutions of the Lagrangian conditions, but they are in the second and fourth quadrants, so have f < 0. At those points, the function f(x,y) does not represent the area of anything and so those two solutions have nothing to do with the original problem. (Remember, Area = 4 |x| |y|, not 4xy.)

    Ask yourself: Is there a positive minimum area the rectangle can have?
     
  12. Nov 15, 2014 #11
    0 area is the smallest the rectangle can have.
     
  13. Nov 15, 2014 #12

    Ray Vickson

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    Right: achieved by "corners" (x,y) = (+-4,0) duplicated and by (x,y) = (0,+-3) duplicated; that is, the rectangle collapses into a line segment, either horizontal or vertical. This is easy to get---no calculus needed at all! For those "solutions" we have either x = 0 or y = 0, and at those points we do NOT have both Lagrangian derivatives = 0. (However, we DO have the more general conditions I listed in my previous post.)
     
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