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Optimization problem

  1. Jan 8, 2004 #1
    Find the area of the largest rectangle with lower base on the x-axis and upper vertices on the curve y = 12-x^2
     
  2. jcsd
  3. Jan 9, 2004 #2
    Since one side is on x-axis then other side will be || x-axis(rectangle) let say y=a then u will x=+sqrt(12-b).

    Therefore u have length say = 2sqrt(12-b) &
    breadth = b.

    Hence Area, [tex]A = b\sqrt{12-b}[/tex].

    u will get b=8 and Length=8& breadth 8.

    Alternate
    Rectangle with max area is a square
    u get [tex] b=2\sqrt{12-b}[/tex]
     
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