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Minimizing amount of fencing

  1. Aug 17, 2015 #1
    1. The problem statement, all variables and given/known data
    Hello!
    It seems that I don't know the correct approach to this problem.

    Sally wants to plant a vegetable garden along the side of her home. She doesn't have any fencing, but wants to keep the size of the garden to 100 square feet.
    a) What are the dimensions of the garden which will minimize
    the amount of fencing she needs to buy?
    b) What is the minimum amount of fencing she needs
    to buy? Round your answers to the nearest foot. (Note: Since one side of the garden will
    border the house, Sally doesn't need fencing along that side.)

    2. Relevant equations
    My thoughts:
    x - width of the garden
    l - length
    Since one wall is not required, the fencing = 2x + l
    area = x*l = 100, hence l = 100/x

    Have no idea how to proceed to solve a) and b) (without calculus, based only on precalculus program).

    Thank you!


    3. The attempt at a solution
     
  2. jcsd
  3. Aug 17, 2015 #2

    Ray Vickson

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    So, you want to minimize the fencing length ##L(x) = 2x + 100/x## over ##x > 0##. One way is to plot a graph of the function and see where the minimum occurs. This would at least give you an approximate estimate of where the minimum occurs.

    Another way (without calculus) is to use the arithmetic/geometric inequality, which says that for ##a, b > 0## we have
    [tex] \frac{a+b}{2} \geq \sqrt{ab}, [/tex]
    with equality holding if, and only if ##a = b##. Sometimes this can give a slick way of minimizing a sum of two terms, such as you have in your problem.
     
  4. Aug 17, 2015 #3

    BvU

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    I Sally forced to lay out a rectangular garden ?
     
  5. Aug 17, 2015 #4

    RUber

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    First put the equation into one variable.
    ##F = 2x + 100/x##
    Then you can multiply by x to get:
    ##2x^2 - Fx + 100 = 0##
    Find possible solutions to the quadratic equation in terms of F, and you will see what the minimum possible F is because of the radical involved. When you have F, work backward to find x and l.
    Like Ray mentioned, a quick plot will at least put you in the ball park, so you know if your solution makes sense.
     
  6. Aug 18, 2015 #5
    Thank you very much for your help!
     
  7. Aug 18, 2015 #6

    BvU

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    well, what did you find ? And is it shorter than a half-circle fence ?
     
  8. Aug 18, 2015 #7

    haruspex

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    You make a good point, but the question does ask for dimensions, plural, and does not ask about shape. So I suspect the questioner intended rectangular.
    On that assumption, there's a way which does not even use algebra, provided you assume that a square would be the ideal rectangle if there were no house wall to make use of.
     
  9. Aug 18, 2015 #8
    1) How many equations do you have? You said: Fence = 2x+l and x*l=100
    2) How many unknowns (or variables) do you have?
     
  10. Aug 18, 2015 #9
    INITIAL GUESS:
    When you have a problem and you do not know how to solve it, one approach is to make an initial guess.

    Example:
    low guess: The width of the garden plot will be 1 foot.
    apply your first formula. Then the length must be l = 100/w = 100 feet
    ... boy does sally have a long house !
    apply your second formula. Then the amount of fence needed must be Fence = 2x+l = 2*1+100 = 102
    put your result in a table
    ... Width Length Fence
    ... 1 100 102

    MAKE A SECOND GUESS:
    Increment: 1 foot
    New width is 2 feet
    Apply the first formula. l = 100/w = 50 feet
    Apply your second formula. Fence = 2x+l = 2*2+50 = 4+50 = 54
    ... Width Length Fence
    ... 1 100 102
    ... 2 50 54

    REVIEW THE RESULT
    Which answer is better?
    Move in the direction that is better.
    In this case, 2 feet is a lot better than 1 feet.
    So now try 3 feet.

    CONTINUE AND REPEAT THE METHOD
     
  11. Aug 21, 2015 #10
    A semi-circle is the logical answer. The problem could hardly have been stated, "What is the dimension of the garden..."
     
  12. Aug 21, 2015 #11

    haruspex

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    No, but it would have asked for shape.
     
  13. Aug 22, 2015 #12
    Only mathematically. Most gardens are rectangular.
     
  14. Aug 22, 2015 #13
    Note that this is not a gardening forum.
     
  15. Aug 22, 2015 #14
    How about the dimensions of a trapezoid (or better yet, a trapezoid next to a rectangle)?
     
  16. Aug 22, 2015 #15

    haruspex

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    I repeat, it does not ask about shape, suggesting the shape is assumed.
     
  17. Aug 22, 2015 #16
    I agree; the OP must assume a shape to get the minimum fence length.
     
  18. Aug 22, 2015 #17

    haruspex

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    No, you are saying the OP must choose a shape. As I'm sure you understand, I am arguing that the problem setter has assumed a specific shape (namely, a rectangle) and has failed to realise it was not made explicit.
     
  19. Aug 23, 2015 #18
    So, the correct answer to the instructor is:

    Assuming you meant a rectangle, it measures 7'X14' requiring 28' of fencing.
    Assuming you meant what you wrote, it's an 8' radius semicircle requiring 25' of fencing.
     
  20. Aug 23, 2015 #19

    Ray Vickson

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    There are two problems with your posting:
    (1) Since the OP has not told us his answer (and we don't know if he did correct work) we should not supply give complete answers for him. PF rules require we give hints only, not answers, at least not until the work has already been turned in for marking, etc.
    (2) Hinting to the instructor his/her failings is likely a bad move on the student's part. Better to leave out the "Assuming you meant..." parts, and just say: "Assuming a rectangular shape..." and/or "Assuming a semi-circular shape ..." . (Even then, proving that the optimal shape is a semi-circle involves constrained calculus of variations, which goes way beyond pre-calculus. Of course, it is intuitive, but intuition is not proof.)
     
  21. Aug 23, 2015 #20
    Sorry, didn't realize PF is a humor-free zone.

    ;-]
     
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