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Math Help - optimization

  1. #1
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    optimization

    Ok, I'm pretty lost on this one.

    A farmer will produce a crop of soybeans between 0 and 250 tons. The price of P of a single ton of soybean is $500 minus twice the number of tons X that are produced, since the unit price lowers as the farmer produces more. In addition, in order to partially control prices, the government pays the farmer an extra $200 for each ton produced. Use calculus to find the number of tons X that the farmer should produce to maximize the revenue received (total dollars earned from both the gov. and from selling X tons at price P).
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  2. #2
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    Quote Originally Posted by kwivo View Post
    A farmer will produce a crop of soybeans between 0 and 250 tons. The price of P of a single ton of soybean is $500 minus twice the number of tons X that are produced, since the unit price lowers as the farmer produces more. In addition, in order to partially control prices, the government pays the farmer an extra $200 for each ton produced. Use calculus to find the number of tons X that the farmer should produce to maximize the revenue received (total dollars earned from both the gov. and from selling X tons at price P).
    Hello,

    let x be the amount of tons of beans
    then
    - the price for 1 t of beans is (500 - 2x)
    - the total revenue is

    r(x) = (500-2x) \cdot x + 200x

    Expand the bracket, collect like terms:

    r(x) = -4x^2+700x

    You'll get the extreme value for r if r'(x) = 0

    r'(x) = -8x+700~with~-8x+700=0~\iff~x=\frac{700}{8}

    Btw: The graph of r is a parabola opening downward which has it's maximum as it's vertex. The x-coordinate of the vertex of a parabola is calculated by

    x_v=-\frac{b}{2a}. With your problem: x_v=-\frac{700}{2\cdot (-4)}
    That means it isn't necessary to use calculus here.
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  3. #3
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    So I don't need to do anything with the information of the first sentence "A farmer...between 0 and 250 tons"? By the way, don't you get -2x^2+700x instead when you expand?

    (500-2x)(x)+200x= 500x-2x^2+200x=-2x^2+700x
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  4. #4
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    Quote Originally Posted by kwivo View Post
    So I don't need to do anything with the information of the first sentence "A farmer...between 0 and 250 tons"? By the way, don't you get -2x^2+700x instead when you expand?

    (500-2x)(x)+200x= 500x-2x^2+200x=-2x^2+700x
    Hi,

    of course you are right (I must have messed my notes so the factor 4 from the derivation popped up in the original equation )

    I don't have any use for the information that the farmer's limits of production are between 0 and 250t because even with the corrected values you'll get a maximum revenue at 175 t which is far below the limit of 250 t.
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  5. #5
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    Thanks. One more question, in case I get another problem like this, I want to know how I would use that information when I get a question that doesn't automatically fall into the given interval.
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  6. #6
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    Quote Originally Posted by kwivo View Post
    Thanks. One more question, in case I get another problem like this, I want to know how I would use that information when I get a question that doesn't automatically fall into the given interval.
    Hello,

    consider your problem starts with the first statement like this:

    A farmer will produce a crop of soybeans between 0 and 150 tons.

    Then you do all necessary calculations and check afterwards if the result fits into the given conditions. Because 175 > 150 the only valid maximum value is the 150. In Germany this is called a boundary extremum.
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