1. ## Maximising a quadratic function.

Here is another one that I am having problems with.

The cost (in dollars) of producing x items is

C (x) = 4000 - 3x + (X^2 1000)

If the items are sold for 4 each, find the value of x that maximises the Profit and find the maximum profit. I have figured out the equasion Profit = P = 4x - (4000- 3x + (X^2 1000) I tried to put it into my graphics calculator, and get the value for the max profit but its a linear equasion. Thanks in advance for any help P.S apologies in advance, I do not know how to put maths equasions into computers... maybe my next question should be about that! 2. You should know that since the x^2 term has a negative coefficient, that there will be a maximum turning point, which you find by completing the square to put the equation into turning point form... \displaystyle \begin{align*}P &=4x - \left(4000 - 3x + \frac{x^2}{1000}\right)\\ &= 4x - 4000 + 3x - \frac{x^2}{1000}\\ &= -\frac{x^2}{1000} + 7x - 4000\\ &= -\frac{1}{1000}\left(x^2 - 7000x + 4\,000\,000\right)\\ &= -\frac{1}{1000}\left[x^2 - 7000x + \left(-3500\right)^2 - \left(-3500\right)^2 + 4\,000\,000\right]\\ &= -\frac{1}{1000}\left[\left(x - 3500\right)^2 - 12\,250\,000 + 4\,000\,000\right]\\ &= -\frac{1}{1000}\left[\left(x - 3500\right)^2 - 8\,250\,000\right]\\ &= -\frac{1}{1000}\left(x - 3500\right)^2 + 8250\end{align*} So what is the maximum profit and how many units do they have to sell? 3. Originally Posted by jlee88 Here is another one that I am having problems with. The cost (in dollars) of producing x items is C (x) = 4000 - 3x + (X^2 1000) If the items are sold for4 each, find the value of x that maximises the Profit and find the maximum profit.

I have figured out the equasion

Profit = P = 4x - (4000- 3x + (X^2 1000)

I tried to put it into my graphics calculator, and get the value for the max profit but its a linear equasion.
Not with that $x^2$ in it!

[quoteThanks in advance for any help

P.S apologies in advance, I do not know how to put maths equasions into computers... maybe my next question should be about that![/QUOTE]