1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Homework Helper

    [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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor
    Homework Helper

    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:
      16 KB
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Applied Min/Max word problem
  1. Absolute Max Min Problem (Replies: 13)