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Math Help - I hate these problems..

  1. #1
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    I hate these problems..

    Ms. Brown has 200 m of fencing with which she intends to construct a rectangular enclosure along a river where no fencing is needed. She plans to divide the enclosure into two parts (as shown) to separate her sheep from her goats.

    What would be the dimensions of the enclosure if its area is to be a maximum? Show a complete algebraic solution.

    I'm lost. =|
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    Quote Originally Posted by xxlvh View Post
    Ms. Brown has 200 m of fencing with which she intends to construct a rectangular enclosure along a river where no fencing is needed. She plans to divide the enclosure into two parts (as shown) to separate her sheep from her goats.

    What would be the dimensions of the enclosure if its area is to be a maximum? Show a complete algebraic solution.

    I'm lost. =|
    ...as shown...?

    where's the diagram?

    does it look like this?

    were there any labels?
    Attached Thumbnails Attached Thumbnails I hate these problems..-fence.jpg  
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    That's the diagram I had drawn out. No labels given, though.
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    Quote Originally Posted by xxlvh View Post
    That's the diagram I had drawn out. No labels given, though.
    this is what we call an optimization problem. there are two things they pivot on (your professor might call them something different, this is just what i've heard them called, students have told me other things). you need a constraint equation and an objective equation. how do we use them?

    start by drawing your diagram and labeling things you will need. then form your constraint equation. it is made up of formulating any restraints you have in the problem. here, for instance, our constraint is the perimeter of the fence. we have 200 meters of fencing. that is restricting us. we must use 200 m, not fewer, not more. so your constraint equation will have to do with the perimeter of the fence. find that equation. we use the constraint equation to solve for one variable in terms of the other.

    your objective equation is a formula for what you want to maximize or minimize, that is, it is the formula through which we obtain our objective to optimize the object in some way. here, it is obviously the formula for the enclosed area of the fence, since that's what they say we must maximize.

    so you have those two things to solve the problem, now look for what you want to find. here we are looking for dimensions. so let me start you off. we will work through this together.

    label all the vertical sides of the fence y, and the one long horizontal side x.

    now, give me the constraint and objective equations
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    3y + x = 200
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    Quote Originally Posted by xxlvh View Post
    3y + x = 200
    yes, that is our constraint. the perimeter is 200 and that is the formula for the perimeter of our fence. here, it is easy to solve for x. so do that, and put that equation aside for now.

    now want is our objective equation?
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    errr. 200 = 3y + (200-x)?
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    Quote Originally Posted by xxlvh View Post
    errr. 200 = 3y + (200-x)?
    ??

    remember, our objective is the area equation. what is the equation for the area of our fence. call the area A, so i want, A = ...?
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    a = xy ?
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    Quote Originally Posted by xxlvh View Post
    a = xy ?
    why do you seem so unsure? yes it's A = xy. that is the equation for the area of a rectangle, and we have a rectangular fence here.

    now working in two variables is not good in pre-algebra, so let's change our objective equation to one variable. we do that using our constraint equation. remember i told you to solve for x and put that equation aside. well, take it back up. i want you to replace x in the area equation with the formula you had x equal to in the constraint equation. and simplify. what do you get?
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    so a=xy and 3y + x = 200 or x=200 - 3y
    so a = (200-3y)(y)
    a = 200y - 3y2
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    Quote Originally Posted by xxlvh View Post
    so a=xy and 3y + x = 200 or x=200 - 3y
    so a = (200-3y)(y)
    a = 200y - 3y2
    good. so we see that we can write the area equation as a quadratic equation. what's more, it is a downward opening parabola. and what do we know about downward opening parabolas? they have a maximum. that is good and convenient. because our objective was to find the dimensions that maximized the area. so since the area can be represented as a downward opening parabola, we can find its maximum by finding the value of y that occurs at the vertex of the parabola.

    there are several ways to find the vertex, and hence, our desired y-value. two of the most predominant ways are: (1) use the vertex formula, and (2) complete the square.

    we shall proceed with which ever method you were taught in class. so what'll be?
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    completing the square is what i'm more used to

    a = -3(y-(100/3))2 - (10 000/9)
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    Quote Originally Posted by xxlvh View Post
    completing the square is what i'm more used to

    a = -3(y-(100/3))2 - (10 000/9)
    apparently you're not used to it enough. try again, that's incorrect
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  15. #15
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    Ahah
    aggghh
    I see -- for some reason I thought 200/3 was negative... so okay, fixed equation. Or at least I hope this is fixed.

    a = -3(y2+(100/3))2 - (10 000/9)
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