Math Question

Discussion in 'Teacher Time Out' started by ku_alum, Mar 21, 2009.

  1. ku_alum

    ku_alum Aficionado

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    Mar 21, 2009

    My husband is building me a raised bed vegetable garden (google square foot gardening if you're intersted). Anyway ... if he uses 8 foot boards and builds it 4 feet by 4 feet we have 16 square feet. If he uses the SAME materials, but builds it 3 feet by 5 feet, we have 15 square feet.

    Am I just having a blond moment? Why does it work that way, why do we lose a foot even though we are using the same materials??
     
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  3. Aliceacc

    Aliceacc Multitudinous

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    Mar 21, 2009

    Because 2 shapes can have the same perimeter but different areas.

    With your 16 linear feet of perimeter, some options include:
    1x7, area 7
    2x6, area 12
    3x5, area 15
    4x4, area 16.

    The square will always maximize the area.
     
  4. ku_alum

    ku_alum Aficionado

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    Mar 21, 2009

    I'm not trying to be difficult ... but, why?
     
  5. Catcherman22

    Catcherman22 Companion

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    Mar 21, 2009

    lets assume each foot holds one plant.. so a 3 X 5 hold 3 plants one direction and 5 plants another direction..

    start with the 4 by 4. 4 plans long and 4 plants wide.

    x x x x
    x x x x
    x x x x
    x x x x

    so to go to a 3 x 5 you need to remove the 4th column (so you have 3 columns).

    x x x x
    x x x x
    x x x x
    x x x x

    To get 5 rows, you need one of each of the removed column and put it on each remaining column.

    x x x
    x x x
    x x x
    x x x
    x x x x

    This leaves the one remaining plant out in the open.. and is hence lost.

    That help?
     
  6. Malcolm

    Malcolm Enthusiast

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    Mar 21, 2009

    A=lw

    4'x4' = 16 square feet

    3'x5' = 15 square feet

    A square will always maximize the area for a given perimeter in a rectangle.

    The more different the side lengths are, the smaller will be the area for the same perimeter.

    2'x6' = 12 square feet

    1'x7' = 7 square feet
     
  7. HMM

    HMM Cohort

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    Mar 25, 2009



    if you have a constraint for the perimeter 2W+2L=C

    and you want to maximize the area A=L*W....you can use a little calculus


    from 2W+2L=C we can get

    W=(C-2L)/2

    plug into A=L*W

    so

    A=L*(C-2L)/2=(1/2)(C*L)-L^2

    findthe derivative of A with respect to L

    A'=C/2-2L

    and set equal to 0 and solve for L

    C/2=2L
    L=C/4

    then W=(C-2L)/2=(C-2C/4)/2=(C/2)/2=C/4


    so L=W

    and this is a max since

    A''=-2<0 for any value of L (2nd derivative test)

    you could have used Lagrange Multiplies if you wanted to but that would have been a little overkill
     
  8. TeacherGroupie

    TeacherGroupie Moderator

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    Mar 25, 2009

    HMM, I'm afraid I could feel ku_alum's eyes go glassy all the way over on the Left Coast...

    Neatly explained, Catcherman.
     
  9. HMM

    HMM Cohort

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    Mar 25, 2009

    Just thought I'd share ;)
     

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