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MATH 11 help PLease!

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A rectangular garden is made at the side of the house. Part of the house forms one side of the fencing and divided into four equal plots. If 90m of fencing is available, determine the dimensions for the entire garden that will produce a maximum area.

how can i come up with a quadratic equation from this problem????

  • MATH 11 help PLease! - ,

    Sketch the problem
    You need 5 lengths of fence of length w and 1 length e for end
    area = A = w e
    5 w + e = 90
    5 w + A/w = 90

    5 w^2 - 90 w + A = 0

  • MATH 11 help PLease! - ,

    where did you get 5?

  • MATH 11 help PLease! - ,

    Now if you do not know calculus, find the vertex of the parabola by completing the square

    5 w^2 - 90 w = -A

    w^2 - 18 w = -A/5

    w^2 -18 w + 81 = -A/5 + 81

    (w-9)^2 = -(1/5)(A - 405)

    so max at w = 9 , A = 405
    then
    5 (9) + e = 90
    45 + e = 90
    e = 45

    so 45 by 9

  • MATH 11 help PLease! - ,

    i know how to get the vertex and all the other stuff,
    i just don't understand where did you get 5?

  • MATH 11 help PLease! - ,

    To get 5, DRAW A Picture

    9 foot fence at 0, 11.25 , 22.5 , 33.75 , 45
    and a 45 foot fence across the end

  • MATH 11 help PLease! - ,

    four fence sections requires five posts :)

  • MATH 11 help PLease! - ,

    because you need one at the start, zero.

  • MATH 11 help PLease! - ,

    (unless you close the figure topologically by joining the last section to the first.)

  • MATH 11 help PLease! - ,

    aah ok.. thanks a lot!!!;)

  • MATH 11 help PLease! - ,

    0, 11.25 , 22.5 , 33.75 , 45
    notice four commas seperating five numbers

  • MATH 11 help PLease! - ,

    I am belaboring this because it is not the last time you will have to think about it.

  • MATH 11 help PLease! - ,

    ok,i will just try it myself so that i can understand it.. thanks..

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