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Finding Max Area of Rectangle in Isos. Triangle

  1. Dec 12, 2007 #1
    1. The problem statement, all variables and given/known data
    An isosceles triangle has base 6 and height 12. Find the maximum possible area of a rectangle that can be placed inside the triangle with one side on the base of the triangle.


    2. Relevant equations
    None.


    3. The attempt at a solution
    Well, so far I've created an X-Y Axis and made the triangle symmetric to the Y-Axis. I am, however, really bad at these and need help from here. Where should I go from here?
     
  2. jcsd
  3. Dec 12, 2007 #2
    I know:

    Area of Rectangle = Max
    A = lw

    I found through the Pythagorean theorem that 3^2 + 12^2 = 12.37 (The length of the hypotenuse)

    I don't know what my secondary equation should be. That's my question.

    As for effort, I've spent 45 minutes on this problem trying to figure it out. I don't know how to get my second equation. Please help!
     
    Last edited: Dec 12, 2007
  4. Dec 12, 2007 #3

    Dick

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    If the rectangle is inside of the triangle there is a linear relation between it's base and height. For example if it's base is 3 then it's height is 6. If base=0 then height=12, if base=6 then height=0. Do you see why? And can you figure it out for any base?
     
    Last edited: Dec 12, 2007
  5. Dec 13, 2007 #4
    Yes, I totally understand that. I got the answer to be 3 by 6 through guess and check, but I need to prove this through calculus.

    How do I get my second equation.

    I have A=L*W, but that's all. I need a second equation so I can get to the max through calculus, not guess and check.

    The linear equation is -2x+12=y
     
  6. Dec 13, 2007 #5
    I got the equation as y= -4x+12 for the right hand slope of the triangle (for x>0)

    and y is also the height of the rectangle at point x, so getting the A=X*Y in terms of X;
    as X=x and Y=-4x+12 ;

    A= x*(-4x+12) which becomes A= -4x^2+12x, then differentiating for 0;
    x=0 or x=1.5

    x cannot = 0 as then the rectangle would have no base width, hence x=1.5.

    subbing x=1.5 into our previous equation (y= -4x+12) to get the height, hence the height is 6.

    But because we only used one half of the isosceles triangle (i.e y= -4x+12 only governs the right hand slope of the triangle ) we must double the x value to get the FULL width which is 2*1.5 = 3.
     
  7. Dec 13, 2007 #6

    Office_Shredder

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    If x is the base, and y the height, then the relationship of x from the midpoint of the triangle to one of its ends and y is y=-4x+12 x from 0 to 3. But the base is in fact twice as long, so the relationship is in fact y=-2x+12 x from 0 to 6. Though I see you doubled at the end to account for the discrepancy
     
  8. Dec 13, 2007 #7

    Dick

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    Ok then. Substitute the linear equation into the area equation A=L*W=x*y=x*(-2x+12). Now use calculus.
     
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