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Thread: Farmer brown problem

  1. #1
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    Farmer brown problem

    Suppose that farmer brown has 260 feet of fencing and wants to build a pen that adjoins the whole side of his 100-ft barn (shown). What should the dimensions be for maximum area?


    Alright, sorry no picture, but essentially we have a perimeter of 260, one of the sides must be at least 100 ft (adjoins the whole barn) and we want maximum area.

    Well, my immediate thought it that a square always has the most area for any rectangle with a given perimeter, so I'm assuming we want what's AS CLOSE to a square as possible, which would be 100X30, yielding an area of 3000 square feet. But I suppose I need to prove this with calculus.

    So.. any suggestions?

    If perimeter is 260:
    $\displaystyle 260 = 2x + 2y$
    Solving for y gives
    $\displaystyle y = 130 - x$

    Okay, y = 130 - x, and y must be at least 100.

    $\displaystyle 100 >= 130 - x$
    $\displaystyle -30 >= -x$
    $\displaystyle 30 <= x$

    So, 100 is greater than or equal to y when 30 is less than or equal to x. So, that gives me an interval of (-infinity, 30] for x.

    $\displaystyle A = xy = x(130-x)$
    $\displaystyle = 130x - x^2$

    $\displaystyle A' = 130 - 2x$

    Okay, so now I need to find all critical points of the derivitive of area, where x =< 30, right?

    so immediately, I have a critical point of 30, because my interval is (-infinity, 30]

    Solve the derivative for 0,

    $\displaystyle 0 = 130 - 2x$
    $\displaystyle 130 = 2x$
    $\displaystyle 65 = x$

    So we have a critical point of 65 (perfect square) and 30, but 65 is not in our interval, therefore the only critical point is 30, therefore 30 is the maximum within the interval, therefore the dimensions for maximum area are 100 x 30.

    Is this adequate proof?

    (EDIT)

    Should I also show that the area is actually increasing on the interval (-infinity, 30] by solving A' for a number within the interval?
    Last edited by OneMileCrash; Jun 30th 2011 at 07:44 AM.
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  2. #2
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    Re: Farmer brown problem

    since there is no "diagram" , I'll have to interpret from your explanation ...




    P = 260 ft of fence + 100 ft of barn = 360 ft

    let x = length of side in addition to the 100 ft barn

    360 = 2(100+x) + 2y

    x + y = 80

    A = (100+x)y = (100+x)(80-x)

    now maximize area
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