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Domain of area function

  1. Sep 16, 2016 #1
    1. The problem statement, all variables and given/known data
    A rectangle has one vertex in quadrant I at the point (x,y) which lies on the graph of y = 2x^2 and another vertex at the point (-x, y) in the second quadrant and the other vertices on the x axis at (-x, 0) and (x, 0)

    What is the domain of the area function?

    y = 2x^2 = w

    l = 2x

    Vertices:

    (-x, 0)
    (x,0)
    (-x,y)
    (x,y)

    2. Relevant equations
    2x(2x^2) = 4x^3

    2x = 2x^2
    3. The attempt at a solution

    Find the zeros:

    0

    Find the maximum area:

    1

    [0,1] is the domain

    But I am not so sure about this. On the one hand a rectangle can't have infinite area. On the other hand, 4x^3, the area function goes all the way to infinity.

    So am I approaching this wrong? How do I find the maximum area if the function has no max? I know that x cannot be any lower than 0 since that would mean negative area and negative area only applies in integral calculus.
     
    Last edited: Sep 16, 2016
  2. jcsd
  3. Sep 16, 2016 #2

    SammyS

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    Please give the complete problem statement.
     
  4. Sep 16, 2016 #3
    I just edited my post to include the whole problem statement.
     
  5. Sep 16, 2016 #4

    SammyS

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    You should have something that looks like a function definition in function notation. Since it represents Area, you might chose A for the function name.

    What is the definition (a formula) for A ?

    A(x) = _?_
     
  6. Sep 16, 2016 #5
    A(x) = 4x^3

    In fact that first equation up there in the second part shows how I got this. 2x is the length and y is the width. Since y = 2x^2, A(x) = 2x(2x^2) = 4x^3
     
    Last edited: Sep 16, 2016
  7. Sep 16, 2016 #6

    SammyS

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    Yes, and you had some other expressions as well .

    Now it's clear that you have the correct function for area.

    Saying that the range of this function goes to infinity, does NOT say that the area can be infinite, It simply means that that area can be made to be as large as desired.
     
  8. Sep 16, 2016 #7

    Ray Vickson

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    You have confused "range" with "domain". The range is the set of area values; the domain is the set of ##x##-values. So, the question is asking about the allowed ##x##-values, not about the values of area.

    Don't worry about the lack of a maximal area. Some functions (on some domains) do not have maxima; this is one of them.

    Of course you cannot have an infinite area, but you can have area greater than 10 million or greater than 100 quadrillion or greater than .... . You see? There is no upper limit beyond which the area cannot go. No matter how large a number you name you can have an area larger than that number.
     
    Last edited: Sep 16, 2016
  9. Sep 16, 2016 #8
    So the domain is [0, ∞)?
     
  10. Sep 16, 2016 #9

    SammyS

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    Can you have x = 0 ?
     
  11. Sep 16, 2016 #10
    Yes. If x = 0, then the rectangle will simply be a point at the origin. Plus, 0 is a valid number for the function.
     
  12. Sep 16, 2016 #11

    SammyS

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    Then it's not a rectangle.

    The point (0,0) is not in Quadrant I .
     
  13. Sep 16, 2016 #12
    But I can have the area function equal 0 and be valid. And the origin is in all quadrants. As x increases, -x decreases by the same amount and y increases by 2x^2 so it traces the parabola as the rectangle gets bigger and bigger. As a result, its area increases by 4x^3.

    I don't see how that changes this:

    [0, ∞)

    to this:

    (0, ∞)
     
  14. Sep 16, 2016 #13

    SammyS

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    The problem statement says what regarding the location of the vertex at (x, y) ?

    Sure the function, f(x) = 4x3 be zero or for that matter it can return a negative, but those are inconsistent with the problem statement as regards A(x) ..
     
  15. Sep 16, 2016 #14
    It just shows that as one of the vertices along with (-x, y), (-x, 0), and (x, 0). It doesn't say anything other than that (x, y) is a vertex as far as the point (x, y) is concerned. The only other thing it shows besides a graph, which I can't put here, is that y = 2x^2 and it asks me these 4 things:

    1) What is the area as a function of x?

    2) What is the domain of the area function?

    3) What is the perimeter as a function of x?

    and

    4) What is the maximum area?

    Everything but the domain, I am confident that I got right.
     
  16. Sep 17, 2016 #15

    SammyS

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    That clearly states that the point (x, y) is in Quadrant I, the first quadrant. That means that x > 0 and y > 0. Neither is equal to zero.
     
  17. Sep 17, 2016 #16

    Ray Vickson

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    If ##x < 0## the length of the side from ##(-x,0)## to ##(x,0)## is ##2|x|##, so whether ##x## is positive or negative, the area is always positive, because ##A = 2 |x| x^2##. The formula ##2 x^3## applies when ##x \geq 0## only. You don't believe it? Well, if ##x = +5## the two ends of the base are at (-5.0) and (5,0), so the distance between them is 10. When ##x = -5## the two ends are at (-5,0) and (5,0) again, so the distance between them is still 10.
     
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