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Algebra based question

  1. Dec 5, 2006 #1
    Hi everyone, I am new to this site. I was wondering if anyone here can help me answer a question, in order for me to study correctly for my math test tomorrow.

    Here's the question:

    Melissa plans to put a fence around her rectangular garden. She has 150 feet of fencing material to make the fence. If there is to be a 10 foot opening left for an entrance on one side of the garden, what dimensions should the garden be for maximum area? This question has to be answered in vertex form.

    If anyone can answer this and explain it step by step to me that would be great!:smile:
     
  2. jcsd
  3. Dec 5, 2006 #2

    Kurdt

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    You have to show some at least some attempt at doing it yourself before anybody can help you.
     
  4. Dec 5, 2006 #3
    This is what I tried so far:
    2L+2W=150ft.- 10ft. / = divide

    2L+2W=140ft.

    2W/2=140ft-2L/2

    W=140ft-L


    A=LW

    A=L(140ft.-L)

    A=140-L*squared

    A(L)=-L*squared+ 140L
     
  5. Dec 6, 2006 #4
    Hold on. Let's go back up here:
    2L+2W=150ft.- 10ft.
    2L+2W=140ft.
    2W/2=140ft-2L/2
    W=140ft-L

    for a second. There is something wrong here. If you divide 1 term by 2, you must divide ________ terms by 2. There is also something else wrong here.

    You are told you originally have 150ft of fencing material. You want to leave a 10ft opening for the door. Does the door require any fencing material? I think you need to look at what the definition of a perimeter is again. Find out what exactly we are doing by going 2L + 2W.
     
    Last edited: Dec 6, 2006
  6. Dec 6, 2006 #5

    Kurdt

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    Well I assume you have to find the largest rectangular area. As KoGs has said there are a few flaws in your original equation. The actual perimeter will be 160 feet as you have 150 feet of fencing plus a 10 foot opening for a drive or whatever.

    You will have:

    [tex] 2L + 2W = 160[/tex]
    [tex] L+W= 80[/tex]

    Now

    [tex] A= LW[/tex]

    So you need to arrange your equation for perimeter to get rid of one of these terms.

    [tex] W=80-L[/tex]

    therefore [tex]A=L\times (80-L) = -L^2+80L [/tex]

    Now you have to solve for the vertex which will maximise the area and the length. I will give you the equation and leave the rest to do for yourself.

    [tex] h = \frac{-b}{2a} [/tex]

    where a & b are the coeficients in the above quadratic equation.
     
  7. Dec 6, 2006 #6
    h=-b/2a is not allowed by my teacher. She said I have to put it in vertex form which I don't know how to.
     
  8. Dec 6, 2006 #7
    Complete the square

    A = 80L - L^2.
    A = -1( L^2 - 80L)
    A= -1 (L^2 - 80L +1600) +1600
    //note it's +1600 at the end because we have to multiply it by -1.

    So we are left with:
    A = -1(L - 40)^2 +1600.
    Now it's clear to see that to maximize A, we have to minimize the -1(L - 40)^2 term. Which of course means L = 40.
    //and in fact this also even gives us what the maximum area is, namely 1600.
    Now we have L, we plug that back into our perimeter formula to get W.

    I'm confused what your teacher means by vector form. Vectors are used to represent positions. But we do not have a single position, what we have is a rectangle.
     
  9. Dec 6, 2006 #8
    haha oh Vertex form.
    Well what points make our Area maximum, and what points make our Area minimum?


    And in fact that is vertex form what I have.

    A = -1( L - 40)^2 + 1600 IS vertext form.
    Does it remind you of something?

    hint: equation of a straight line.
    y = mx + b.

    Now what does an equation of a parabola look like?
     
    Last edited: Dec 6, 2006
  10. Dec 6, 2006 #9
    So the answer would be the dimensions equal 40by40

    A parabola would include the axis of symmetry and the vertex right?
     
  11. Dec 6, 2006 #10
    Pretty much, the standard (vertex) form of a parabola is like the slope intercept form of a line
    y-k = a(x-h)^2 where (h,k) is the vertex. The x coordinate of the vertex in a standard parabola IS the axis of symmetry

    It somewhat resembles a circle equation, too. You can graph a parabola easily just by looking at the standard form
    (x-h)^2 + (y-k)^2 = r^2 (equation of a circle)
     
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