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Math Help - Calc - Word problem

  1. #1
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    Calc - Word problem

    A special CD is manufactured and its manufacturing costs is given by the equation C(x) = 0.08x^2 + 0.2x + 3200. If the price function is given by the equation p(x) = 33 - 0.02x, determine the price that should be charged

    a) to maximize profits

    P(x) = R(x) - C(x)

    R(x) = \frac {33 - 0.02x} {x}

    R(x) = \frac {33}{x} - 0.02

    P(x) = \frac {33}{x} - 0.02 - 0.08x^2 - 0.2x - 3200

    P(x) = \frac {33}{x} -3200.02 - 0.2x - 0.08x^2

    P'(x) = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2

    0 = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2

    3200.02 = \frac {33}{x^2} - 0.2x - 0.08x^2

    3200.02 = \frac {33 - 0.2x^{3} - 0.008^{4}}{x^2}

    3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}

    3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}

    Now I don't know what to do. . .? I can't seem to find the zeros of the first derivative. . .
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  2. #2
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    What does x represent?
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  3. #3
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    Let x represent the number of CDs to maximize profit
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Macleef View Post
    A special CD is manufactured and its manufacturing costs is given by the equation C(x) = 0.08x^2 + 0.2x + 3200. If the price function is given by the equation p(x) = 33 - 0.02x, determine the price that should be charged

    a) to maximize profits

    P(x) = R(x) - C(x)

    R(x) = \frac {33 - 0.02x} {x}

    R(x) = \frac {33}{x} - 0.02

    P(x) = \frac {33}{x} - 0.02 - 0.08x^2 - 0.2x - 3200

    P(x) = \frac {33}{x} -3200.02 - 0.2x - 0.08x^2

    P'(x) = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2

    0 = \frac {33}{x^2} -3200.02 - 0.2x - 0.08x^2

    3200.02 = \frac {33}{x^2} - 0.2x - 0.08x^2

    3200.02 = \frac {33 - 0.2x^{3} - 0.008^{4}}{x^2}

    3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}

    3200.02x^2 = 33 - 0.2x^{3} - 0.008^{4}

    Now I don't know what to do. . .? I can't seem to find the zeros of the first derivative. . .
    You differentiated wrong...I'll help you through this.

    P(x)= \frac {33}{x} -3200.02 - 0.2x - 0.08x^2

    P'(x)=-\frac{33}{x^2}-0.2-0.16x

    Setting equal to zero, we get:

    0=33+0.2x^2+0.16x^3 (I multiplied through by x^2)

    I would recommend using your calculator to find the values of x

    I get x\approx -6.63.

    This doesn't seem right. Did you used the correct formula for revenue? Shouldn't it be R(x)=x\cdot P(x)??

    When R(x)=x\cdot P(x), I get x\approx 59 CDs.

    This is more reasonable.
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  5. #5
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    Yes you are correct, revenue is x(px) ... I was thinking of the average cost formual -_-

    and when I took the derivative of the new formula, I get 164 CDs
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Macleef View Post
    Yes you are correct, revenue is x(px) ... I was thinking of the average cost formual -_-

    and when I took the derivative of the new formula, I get 164 CDs
    What did you get for the new profit function? I get P(x)=-0.28x^2+32.8x-3200. Thus, P'(x)=-0.56x+32.8.

    Setting equal to zero, I get x=\frac{32.8}{.56}\approx 58.57 \implies 59 CDs.
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