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Triangle in 1st quadrant

  1. Sep 8, 2009 #1
    1. The problem statement, all variables and given/known data
    a. The area of a triangle is formed in the first quadrant by the x and y axes and a line through the point (2,1).
    i. Write the area of the triangle as a function of x.
    ii.Determine the domain of the function.
    iii.Is there a maximum and minumum area for some value of x? If so, find the value analytically and graphically.

    3. The attempt at a solution
    I did part i and ii and only really need help with iii.
    For i, I drew a graph of the first quadrant with a point at (2,1) and an arbitrary line through it and both axes. I have attached a graph from paint. I found the slope of the blue line (-1/x-2) and set it equal to the slope of the whole line (-y/x) in order to solve for y in terms of x. Because y= x/(x-2), the area of the triangle is:

    A= 1/2 bh
    =1/2 xy
    = (x^2)/2(x-2)

    ii. D: (2,infinity)

    iii. I know there must be a maximum and minimum area as x approaches 2 and infinity, but I don't know how to "find the value analytically and graphically".
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 8, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Well, for local extrema, what can you say about dA/dx?.... Is there anywhere else that global extrema may occur?
     
  4. Sep 9, 2009 #3
    Ok, you've done i) and ii) well.

    Now, use
    [tex]\begin{array}{l} (\sqrt{a} - \sqrt{b}) ^ 2 \geq 0 \\ \Rightarrow (\sqrt{a}) ^ 2 - 2\sqrt{a}\sqrt{b} + (\sqrt{b}) ^ 2 \geq 0 \\ \Rightarrow a - 2\sqrt{ab} + b \geq 0 \\ \Rightarrow a + b \geq 2 \sqrt{ab} \end{array} [/tex]

    But first...

    [tex]A=\frac{x^2}{2(x-2)}[/tex]

    [tex]A=\frac{x^2-4+4}{2(x-2)}[/tex]

    Now, what can you do?
     
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