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Optimization of a Triangles Area.

  1. Apr 12, 2012 #1
    Hello everyone,

    This is my first post on here, I was hoping it wouldn't be asking for help but I don't have any options left and it's a problem that is due soon. I promise I have tried to solve it myself but I'm unsure if I'm doing it correctly and it is part of a take home test.

    1. The problem statement, all variables and given/known data
    The question is:
    You are planning to close off a corner of the first quadrant with a line segment 19 units long running from (x,0) to (0,y). Show that the area of the triangle enclosed by the segment is largest when x=y.



    2. Relevant equations
    I have A=1/2bh (area = 1/2 base * height) and I know that is what I have to take the derivative of and I know I have to set that equal to zero and solve for it.


    What I'm not sure is if I can consider this a right triangle. It says the "a corner of the first quadrant" so I'm assuming that this is the first quadrant of the Cartesian graph and that it is indeed a right angle.

    3. The attempt at a solution
    I've tried using the Pythagorean theorem, solving for one of the legs (e.i. b = (19^2 - h^2)^1/2 [ b = base and h= height]) and I put that into the area equation, simplify, and take the derivative of it.

    I think I keep screwing the derivative up while doing that. I'm also not sure if I can even be using the Pythagorean theorem.
    I've asked other people in my class, I've checked three textbooks looking for similar examples, I've search for videos like it, search the web and I can't find anything like this to help me.

    Any help would be greatly appreciated.
     
    Last edited: Apr 12, 2012
  2. jcsd
  3. Apr 12, 2012 #2

    cepheid

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    Yes, of course it's a right triangle. Two of its sides are formed by the x and y axes, which are perpendicular to each other.

    Since it's a right triangle, you can use Pythagoras, and therefore x^2 + y^2 = 19^2.

    Can you post the work you did in trying to differentiate the expression for the area, so that we can see whether you are indeed "screwing it up?"
     
  4. Apr 12, 2012 #3
    Thanks Cepheid!

    Since posting that I've been trying and retrying to get the answer and I think I've actually got the right answer now but I'd love it if someone could confirm it, so I will write it out.

    Primary equation:
    A = 1/2bh

    Secondary equation:
    b^2 + h2=192
    Solved for b:
    b = (192 - h2)1/2


    Plugging the secondary into the primary:
    A=1/2h(192 - h2)1/2

    Taking the derivative:
    1/2(192 - h2)1/2 + 1/2h ( 1/2(192 - h2)-1/2 * (-2h)
    Combining the terms and factoring out a 1/2 gives me
    1/2 [ (192 - h2)1/2 + (-h2(192 - h2)-1/2)]

    I gave both terms the same denominator of (192 - h2)1/2

    This gives me
    (192 - h2)1/2*(192 - h2)1/2 -h2
    in the numerator and a denominator of (192 - h2)1/2

    I then multiplied the two square roots in the numerator together to get
    192 - h2 - h2
    in my numerator (denominator is still (192 - h2)1/2 )

    Then I set this equal to 0
    multiply both sides by
    (192 - h2)1/2
    That leaves me with
    192 - h2 - h2 = 0

    Then I solve for h
    192 = 2h2
    361 = 2h2
    361/2 = h2
    ±(361/2)1/2 = h

    This gives me 19/√2
    then (19√2)/2

    I think that is my answer. It's approximately 13.4 and what I plug that back into
    b = (192 - h2)1/2
    I get approximately 13.4 for b

    I know this is a lot to read through but does this seem like the correct answer?
    I hope so because it's 2am here and I'm tired of doing the derivative on this. ;)
     
    Last edited: Apr 12, 2012
  5. Apr 12, 2012 #4

    cepheid

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    I didn't read through all the math, but recall that you were supposed to show that the area was largest when x = y (or, in your notation, b = h). If x = y, then 2x^2 = 19^2, and hence x = 19/sqrt(2) = 13.43.

    So it would seem that you arrived at the result that you were asked to show.
     
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