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Homework Help: Applied Min/Max word problem

  1. Nov 10, 2005 #1
    I can't seem to figure out this problem.
    Find the dimensions of the rectangle with maximum area that can be inscribed in a circle of radius 10.
    I start by drawing the diagram and it seems to me like the circle radius corresponds with a line from the center of the rectangle to one of the rectangle points on the edge of the circle. This could give me a triangle with pythagorean theorem. I should find the value of x that yields largest area as my priority, after that I can find y easily.

    Latex is not behaving today so i'll try my best without. y/2 = (10^2-(x/2)^2)^(1/2)

    Area of a square is xy, duh.

    Plugging this in gives me a chain rule problem which ultimately comes out with the critical points 0, -10 and 10, all of which yield 0 Area (obviously), there has to be a critical point that I am missing between 0 and 10 but the derivative of A(x) doesn't yield any.

    I'm stumped.
  2. jcsd
  3. Nov 10, 2005 #2


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    Try writing the area of the inscribed rectangle as a function of the angle between its diagonals.
  4. Nov 10, 2005 #3


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    You can do that but I'm not sure that is the best way.

    The circle is, of course, x2+ y2= 100.

    The area of the rectangle is (2x)(2y)= 4xy.

    [tex]y= \sqrt{100- x^2}[/tex]
    so the area is [tex]A= 4x\sqrt{100- x^2}= 4x(100-x^2)^{\frac{1}{2}}[/tex].
    Differentiating that and setting the derivative equal to 0 gives a quadratic equation so I'm not at all clear how you got three "critical points"!
  5. Nov 10, 2005 #4
    That's about where i'd got to, I just overcomplicated things.
    the derivative I get is
    [tex] 4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}} [/tex]

    EDIT: Wait a sec... I had the derivative slightly wrong...
    Last edited: Nov 10, 2005
  6. Nov 10, 2005 #5
    Right, I had the derivative wrong, I still came out with -10, 0 and 10 but the +/-10 were asymptotes according to table and x=0 came out as 40 max area.

    That makes absolutely no sense.
  7. Nov 10, 2005 #6
    I get area of the rectangle to be x * Square root (100 - x ^2)

    Still shouldn't make any difference, at least as regards to the derivative of area being equal to zero.

    You might want to look a little closer at how you used the chain rule on this..
  8. Nov 10, 2005 #7


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    [tex] 4(100-x^2)^{\frac{1}{2}} - 4x^2(100-x^2)^{-\frac{1}{2}} [/tex]

    [tex]\frac{4(100-x^2)}{(100-x^2)^{\frac{1}{2}}} - \frac{4x^2}{(100-x^2)^{\frac{1}{2}}} = 0[/tex]


    [tex]4(100-x^2) - 4x^2 = 0[/tex]
    [tex]400 - 8x^2 = 0[/tex]
    [tex]x^2 = 50[/tex]

    only one solution for x.
  9. Nov 10, 2005 #8


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    Perhaps not but [itex]A = 50 \sin \theta[/itex] looks simpler! :smile:
  10. Nov 10, 2005 #9


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    Personally, I like the "it must be a square, by symetry" approach.
  11. Dec 26, 2007 #10

    (The diagram has been attached)

    Area of rectangle = l X b = 2x√(10-x²)

    Squaring the function and discarding the constant

    A = x²(10-x²)
    A = 10x² - x4

    dA = 20x - 4x³

    For turning point dA/dx = 0
    20x - 4x³ = 0
    x = ±√5 , 0

    x = +√5 as length cannot be negative or zero

    d²A = 20 - 12x²

    d²A/dx² < 0 (Maximum)

    A = 2x√(10 - x²)
    A = 10v5 sq. units

    length = 2√(10 - x²) = 2√5
    width = 2x = 2√5

    (Syed Ahsan Badruddin) ©

    Attached Files:

    • Q.JPG
      File size:
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