1. ## Help

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?

2. 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

3. 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,
$\displaystyle \frac{-b}{2a}$, in this case we have,
$\displaystyle \frac{-60}{-4}=15$. Substitute that into the equation to get,
$\displaystyle -2(15)^2+60(15)-25=425$
Q.E.D.