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Math Help - Optimization

  1. #1
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    Optimization

    1. The company's cost C(x)=x^2-3x+64 dollars to produce x items The selling price p when x hundred units are produced is p(x)=1/4(44-x). Determine the level of production (# of items produced) that maximizes profit.
    Here are my steps:

    R(x)= x(1/4)(44-x)=11x-(1/4)x^2
    R(x)=C(x)
    11x-(1/4)^2=x^2-3x+64
    derive: 11-(1/2)(x)=2x-3
    -(1/2)x-2x=-14
    -5/2(x)=-14
    5x=28
    x=28/5=5.6
    Since Profit is P(x)= R(x)-C(x):
    P(5.6)=(5.6)(1/4(44-5.6))-5.6^2+3(5.6)-64
    I got -24.8 which i know is not the right answer. Please help.
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  2. #2
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    Quote Originally Posted by driver327 View Post
    1. The company's cost C(x)=x^2-3x+64 dollars to produce x items The selling price p when x hundred units are produced is p(x)=1/4(44-x). Determine the level of production (# of items produced) that maximizes profit.
    Here are my steps:

    R(x)= x(1/4)(44-x)=11x-(1/4)x^2
    R(x)=C(x)
    11x-(1/4)^2=x^2-3x+64
    derive: 11-(1/2)(x)=2x-3
    -(1/2)x-2x=-14
    -5/2(x)=-14
    5x=28
    x=28/5=5.6
    Since Profit is P(x)= R(x)-C(x):
    P(5.6)=(5.6)(1/4(44-5.6))-5.6^2+3(5.6)-64
    I got -24.8 which i know is not the right answer. Please help.
    The profit is P(x)=R(x)-C(x). You need to maximise the profit, so
    you will be looking for the solutions of:

    P'(x) = R'(x)-C'(x) = 0.

    Then you will have to confirm that this solution is indeed a maximum.

    RonL
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  3. #3
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    You solution is correct. As the graph shows the is something wrong with the statement.
    Attached Thumbnails Attached Thumbnails Optimization-prfit.gif  
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  4. #4
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    Quote Originally Posted by driver327 View Post
    1. The company's cost C(x)=x^2-3x+64 dollars to produce x items The selling price p when x hundred units are produced is p(x)=1/4(44-x). Determine the level of production (# of items produced) that maximizes profit.
    Here are my steps:

    R(x)= x(1/4)(44-x)=11x-(1/4)x^2
    It says the the price is 1/4(44-x) for x hundred units. So shouldn't it be

    R(x) = x(1/4)(44-x/100)=11x-(1/400)x^2?
    R(x)=C(x)
    This is not true. There is no reason R(x) = C(x). To maximize profits, set

    P'(x) = R'(x) - C'(x) = 0.

    This is what you end up with, but it does not come from R(x) = C(x).

    11x-(1/4)^2=x^2-3x+64
    derive: 11-(1/2)(x)=2x-3
    -(1/2)x-2x=-14
    -5/2(x)=-14
    5x=28
    x=28/5=5.6
    Since Profit is P(x)= R(x)-C(x):
    P(5.6)=(5.6)(1/4(44-5.6))-5.6^2+3(5.6)-64
    I got -24.8 which i know is not the right answer. Please help.
    What is the right answer? I get this.

    Setting P'(x) = 11 - 2/400x - 2x + 3 = 0,
    -(1/200+2)x = -14,
    x = 2800/401 = 6.98.
    P(x) = 11x-(1/400)x^2 - x^2+3x- 64 = -15.12.
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  5. #5
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    JakeD
    I think that the poster simply had a typo.
    It is true that maximum profit occurs if marginal revenue equals marginal cost: R=C & R<C.
    I think that is what was meant, even if it was misrepresented.

    I also think that there are errors in your calculations. I get the same answers a were posted.
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  6. #6
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    ok, I used the P'=R'-C'=0.
    (44x-x^2/4)-(x^2-3x+64)
    derive R'= (-(4*(44-2x))/4^2)-(2x-3)=0 Because the 4 becomes a zero.
    take out 2 from num and denom.
    (-(2*(22-x)/8)-(2x-3)=0
    (-(2*(22-x)/8)=2x-3
    -(2*(22-x)=16x-24
    -44+2x=16x-24
    -44=14x-24
    -20=14x
    -10/7=x
    -1.4285...=x
    What else did I do wrong?

    Also, I double read it for typos, That is what it said word for word.
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  7. #7
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    Quote Originally Posted by driver327 View Post
    1. The company's cost C(x)=x^2-3x+64 dollars to produce x items The selling price p when x hundred units are produced is p(x)=1/4(44-x).
    Quote Originally Posted by JakeD View Post
    It says the the price is 1/4(44-x) for x hundred units. So shouldn't it be

    R(x) = x(1/4)(44-x/100)=11x-(1/400)x^2?

    What is the right answer? I get this.

    Setting P'(x) = 11 - 2/400x - 2x + 3 = 0,
    -(1/200+2)x = -14,
    x = 2800/401 = 6.98.
    P(x) = 11x-(1/400)x^2 - x^2+3x- 64 = -15.12.
    Quote Originally Posted by Plato View Post
    JakeD
    I think that the poster simply had a typo.
    It is true that maximum profit occurs if marginal revenue equals marginal cost: R’=C’ & R”<C”.
    I think that is what was meant, even if it was misrepresented.

    I also think that there are errors in your calculations. I get the same answers a were posted.
    Plato,

    The problem statement says cost C(x) is for x items while price p(x) is for x hundred units. So the price for x units would be

    p(x/100) = (1/4)(44-x/100)

    and revenue would be

    R(x) = xp(x/100) = x(1/4)(44-x/100)=11x-(1/400)x^2.

    I solved the problem under this assumption, so I got a different answer than you and driver327.
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  8. #8
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    I found the value for x. 68/18=x=3.77778. mult. by a 100, i got 378 roughly. I plugged 378 into P(378)=R(378)-C(378)=-173377.
    R((378)=(378(1/4)(44-378)-C=(378)^2-3(378)+64 then I got the answer -173377. If this is not right, I officially give up.
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