# Help

• Mar 5th 2006, 07:52 AM
batman123
Help
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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?
• Mar 5th 2006, 08:56 AM
topsquark
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
• Mar 5th 2006, 10:05 AM
ThePerfectHacker
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,
$\displaystyle \frac{-b}{2a}$, 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.