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Math Help - Writing Algebraic Equations(Help Me)

  1. #1
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    Writing Algebraic Equations(Help Me)

    An apple grower finds that an apple tree produces on an average 400 apples per years, if no more than 16 trees are planted in a limit area. For each additional tree planted per unit area, the grower finds that the yield decreases by 20 apples per tree per years. How many trees should the grower plant per unit area so as to get the maximum yield?

    Please teach me how to form the equitions... don't just give me answer... Thanks
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  2. #2
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    any1 help me plz?
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  3. #3
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    Quote Originally Posted by huemunyeow View Post
    An apple grower finds that an apple tree produces on an average 400 apples per years, if no more than 16 trees are planted in a limit area. For each additional tree planted per unit area, the grower finds that the yield decreases by 20 apples per tree per years. How many trees should the grower plant per unit area so as to get the maximum yield?

    Please teach me how to form the equitions... don't just give me answer... Thanks
    yield = trees x apples

    trees = (16 + x) since you start with 16 and add x trees.

    apples = (400 - 20x) because each tree lose 20 apples from the initial 400 for every new tree x.

    yield = (16 + x)(400 - 20x) = -20x^2+80x+6400

    I assume you don't know calculus, so you graph the yield function and you will see that it has a peak at x = 2. Therefore the grower should plant 16+2=18 trees. His yield is then -20(2)^2+80(2)+6400=-80+160+6400=6460 apples.
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  4. #4
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    Quote Originally Posted by xifentoozlerix View Post
    yield = trees x apples

    trees = (16 + x) since you start with 16 and add x trees.

    apples = (400 - 20x) because each tree lose 20 apples from the initial 400 for every new tree x.

    yield = (16 + x)(400 - 20x) = -20x^2+80x+6400

    I assume you don't know calculus, so you graph the yield function and you will see that it has a peak at x = 2. Therefore the grower should plant 16+2=18 trees. His yield is then -20(2)^2+80(2)+6400=-80+160+6400=6460 apples.
    I appreciate what u had done for my question. But, according to my teacher, calculus is used to find the maximum yield. therefore, i need a cubical (X^3) equation in order to use d^2y/dx^2 to find the maximum yield.

    I am going to have an exam tomorrow.... there should be a same type of question in the exam paper, if some1 know how to form an equation out of this, please spend a little time on the question. Thanks
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  5. #5
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    Quote Originally Posted by huemunyeow View Post
    I appreciate what u had done for my question. But, according to my teacher, calculus is used to find the maximum yield. therefore, i need a cubical (X^3) equation in order to use d^2y/dx^2 to find the maximum yield.

    I am going to have an exam tomorrow.... there should be a same type of question in the exam paper, if some1 know how to form an equation out of this, please spend a little time on the question. Thanks
    The equation for the yield is quadratic, not cubic, and you don't need a second derivative to find extrema. In this case, the second derivative is the constant -40, which guarantees that there is only one critical point and it will be an absolute maximum.

    yield=-20x^2+80x+6400

    \frac{d(yield)}{dx}=-40x+80=0 \implies 80=40x \implies x=2

    x=2 is then a critical point. Again, it is the only critical point. You can tell it is a maximum without the second derivative test because if you plug in x=1 into -40x+80, you get a positive number, and when you plug in x=3 into -40x+80, you get a negative number. This means the function is increasing before 2 and decreasing after 2. Once again, this means the farmer should grow 16+2=18 trees per unit area.
    Last edited by xifentoozlerix; February 20th 2008 at 12:01 AM.
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  6. #6
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    i had solved the question. Thanks man.. about the cubic equation, it was my friend's infomation and it was not my teacher's information. he had lead me to somewhere else and made me thinking about this question for 2 days. OMG
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