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    Michele a business consultant has developed a management training course. She has determined that her yearly profit in hundreds of dollars from selling and conducting the course is given by P(x)=-2x^2+60x-25. How many courses should she sell to maximize her profit? what is the maximum profit?
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    Quote Originally Posted by batman123

    Michele a business consultant has developed a management training course. She has determined that her yearly profit in hundreds of dollars from selling and conducting the course is given by P(x)=-2x^2+60x-25. How many courses should she sell to maximize her profit? what is the maximum profit?
    The maximum of a function (or minimum) occurs when the first derivative is equal to zero. So find x such that P'(x)=0. That this will be a maximum is due to the orientation of the quadratic. (Or you can do the second derivative test.) Then use that value of x in the function P to find the maximum profit.

    -Dan
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    Quote Originally Posted by batman123

    Michele a business consultant has developed a management training course. She has determined that her yearly profit in hundreds of dollars from selling and conducting the course is given by P(x)=-2x^2+60x-25. How many courses should she sell to maximize her profit? what is the maximum profit?
    This is a parabola with a maximum (notice the negative number in front). The max/min of a parabola is given by,
    \frac{-b}{2a}, in this case we have,
    \frac{-60}{-4}=15. Substitute that into the equation to get,
    -2(15)^2+60(15)-25=425
    Q.E.D.
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