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Math Help - Quadratic Function problem

  1. #1
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    Quadratic Function problem

    A company sells X units of a product whose profits function is given by the quadratic model P(X) =60X-X^2 = X(60-X)

    A) How many units should the company sell in order to maximize their profits

    B) What will be the maximum profit if the # of units found in part (A) were sold

    C) what is the maximum number of units the business can sell and still make a pofit

    Once again i am stopped by quadratic functions please help me
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by waterboy View Post
    A company sells X units of a product whose profits function is given by the quadratic model P(X) =60X-X^2 = X(60-X)

    A) How many units should the company sell in order to maximize their profits

    B) What will be the maximum profit if the # of units found in part (A) were sold

    C) what is the maximum number of units the business can sell and still make a pofit

    Once again i am stopped by quadratic functions please help me
    Using P(x)=ax^2+bx+c, the maximum point is reached when x=\frac{-b}{2a}

    So, in P(x)=-x^2+60x \;,x=\frac{-60}{-2}=30.

    Profit would be P(30)=900.
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  3. #3
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    Thank you masters but can you show me how you rearranged THE PROBLEM

    60X-X^2=X(60-X)

    i DID THIS

    60X-X^2=60X-X
    X^2=X^2

    60X=X^2+60X-X

    where did you get a,b and c
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by waterboy View Post
    Thank you masters but can you show me how you rearranged THE PROBLEM

    60X-X^2=X(60-X)

    i DID THIS

    60X-X^2=60X-X
    X^2=X^2

    60X=X^2+60X-X

    where did you get a,b and c
    The standard form of a quadratic function is:

    f(x)=ax^2+bx+c=0

    a is the coefficient of x^2 and b is the coefficient of x.

    Your function was P(x)=60x-x^2

    If you rearrange your function into standard form, it would look like this:

    P(x)=-x^2+60x, where a = -1, b = 60.

    This is the graph of a parabola that opens downward. The vertex would be the maximum point on the parabola. The x-coordinate of the vertex is given by the formula:

    x=\frac{-b}{2a}

    This formula is derived by finding the x-intercepts and dividing by 2, but that's another story. Just remember this formula.

    Substituting the coefficients into that formula, I found the x-coordinate to be 30.

    Substituting P(30) into P(x)=-x^2+60x, I found the y-coordinate of the vertex, which is the maximum value (profit) when 30 units are sold. P(30) = 900.
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