1. ## Profit function!

P(X) = -60X2 + 1500X - 4000 - 1 x 25

This is the profit function and its domain,

1) the value of X where there is a breakeven point
2) the value of X where profit be maximum

thankyou so very much!!!

2. Originally Posted by karimwahab
P(X) = -60X2 + 1500X - 4000 - 1 x 25

This is the profit function and its domain,

1) the value of X where there is a breakeven point
2) the value of X where profit be maximum

thankyou so very much!!!
The breakeven point is the point where there is no profit and no deficit. In other words, the poitns when $\displaystyle P(x) = 0$. You find x values of this by finding the roots of the equation. Using the quadratic formula!

$\displaystyle P(x) = -60x^2+1500x-4000 = 0$

$\displaystyle x = \frac{-1500 \pm \sqrt{1500^2-4(-60)(-4000)}}{2(-60)}$

$\displaystyle x = \frac{-1500 \pm \sqrt{1290000}}{-120}$

$\displaystyle x = 3.035$ or $\displaystyle x = 21.964$

For the maximum, the first derivative of the function is 0. Solve for x, and then plug that x value into the original equation to find P(x) for that value of x. The 2nd derivative test will tell you whether you have a maximum or a minimum. (Although you will have a maximum, since your coefficients of x^2 is negative!)

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# what is domain for profit

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