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Word Problem

  1. Jun 12, 2005 #1
    Hello I have this word problem that I am having problems solving. Hopefully someone can help

    Here is it.

    A piece of wire 10m long is cut into two pieces. On piece, of length x, is bent into the shape of a square. The other is bent into the shape of a equilateral triangle.

    (a) expess the total area of a enclosed as a function of x
    b. For what value is of x is this total area a minimum.

    This is what I think I can figure out.

    One side of the square is s = x/4
    One side of the triangle is s = (10 -x)/3

    Area of Square is A= lw
    Area of Triangle is A = 1/2bh

    How do I go from here.

    TIA

    I may be totally wrong
     
  2. jcsd
  3. Jun 12, 2005 #2

    arildno

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    You're on the right track!
    Let's take the area of the square first.
    Since the length is equal to the width for the square, we have that the square's area is:
    [tex]A_{sq.}=(\frac{x}{4})*(\frac{x}{4})=\frac{x^{2}}{16}[/tex]

    Agreed?

    Now, you've got an equilateral triangle, whose side is [tex]s=\frac{10-x}{3}[/tex]
    What you need to figure out, is the height of that triangle.
    To do that, look on the triangle formed by dropping a perpendicular from one vertex down on one side (dividing that side in half), whose side-lengths must be [tex]s,\frac{s}{2},h[/tex] where "h" is the height.
    Our triangle is right-angled, so we have, by Pythogoras:
    [tex]s^{2}=(\frac{s}{2})^{2}+h^{2}[/tex]
    that is we get:
    [tex]h=\frac{\sqrt{3}}{2}s[/tex]

    Now, can you take it from here?
     
    Last edited: Jun 12, 2005
  4. Jun 12, 2005 #3
    how did you get

    [tex]h=\frac{\sqrt{3}}{2}s[/tex]


    from

    [tex]s^{2}=(\frac{s}{2})^{2}+h^{2}[/tex]


    ??
     
  5. Jun 12, 2005 #4

    arildno

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    [tex]h^{2}=s^{2}-\frac{s^{2}}{4}=\frac{4s^{2}-s^{2}}{4}=\frac{3}{4}s^{2}\to{h}=\frac{\sqrt{3}}{2}s[/tex]
     
  6. Jun 12, 2005 #5
    Yeah I should have seen that.

    So now that I know that the height

    [tex]h=\frac{\sqrt{3}}{2}s[/tex]


    I can plug s into the height formula and the height formula into the area like this

    [tex]A=\frac{1}{2}(\frac{10-x}{6})(\frac{\sqrt{3}}{2})(\frac{10-x}{3})[/tex]

    [tex]A=\frac{100\sqrt{3}-20x\sqrt{3}-x^2\sqrt{3}}{72}[/tex]

    Now I add the to area formulas togther to get

    [tex]A=\frac{100\sqrt{3}-20x\sqrt{3}-x^2\sqrt{3}}{72} + \frac{x^2}{4}[/tex]

    Is this correct?
     
  7. Jun 12, 2005 #6

    arildno

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    Basically, yes,; but you've made a couple of arithmetical mistakes.

    You should have for the total area A:
    [tex]A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}[/tex]
     
  8. Jun 12, 2005 #7
    cool thank i will try to figure out where made the mistakes.

    how do i figure out this part

    For what value is of x is this total area a minimum

    Thanks again
     
  9. Jun 12, 2005 #8

    arildno

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    Have you learnt about differentiation (derivatives) yet?
     
  10. Jun 12, 2005 #9
    no not yet
     
  11. Jun 12, 2005 #10

    arildno

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    And you were given this as an assignment all the same??

    Are you sure about that?
    Haven't you covered minimization problems in class?
     
  12. Jun 12, 2005 #11
    in this

    [tex]A=(\frac{x}{4})^{2}+\frac{\sqrt{3}}{4}(\frac{10-x}{3})^{2}[/tex]

    what happen 1/2 from the area of triangle??

    Just wondering.
     
  13. Jun 12, 2005 #12
    The half has already been taken into account, that's why you have [tex]\frac{\sqrt{3}}{4}[/tex] and not [tex]\frac{\sqrt{3}}{2}[/tex], ie [tex]\frac{\sqrt{3}}{2}\frac{1}{2} = \frac{\sqrt{3}}{4}[/tex].
     
    Last edited: Jun 12, 2005
  14. Jun 12, 2005 #13

    arildno

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    You have, for the area of the triangle:
    [tex]A_{tr}=\frac{1}{2}*h*s=\frac{1}{2}*\frac{\sqrt{3}}{2}*s^{2}=\frac{\sqrt{3}}{4}s^{2},s=\frac{10-x}{3}[/tex]
     
  15. Jun 12, 2005 #14
    all we have done with min or max is in the for of ax^2 + bx + c

    if a < 0 then x= -(b)/2a is the max
    if a > 0 then x= -(b)/2a is the min

    this was a assignment and I got it wrong and think there will be something like it on the test. This is the only min max question we have had.

    thanks for all your help
     
  16. Jun 12, 2005 #15
    I think I need to be a little more organized when I am solving a question so I know what I have done already. Thanks for the help.
     
  17. Jun 12, 2005 #16

    arildno

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    Yes, then what you need to do is to identify "a" and "b" in your expression for A(x)!
     
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