Thread: 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!

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...



So what is the maximum profit and how many units do they have to sell?

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 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.

Not with that [itex]x^2[/itex] 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]