1. ## Basic linear, advice needed

Hi guys, I'm think im missing a step here:

cost in pounds = C=500+10X *(x=no. units produced)
revenue in pounds = R=150X - 3X^2

I work out the weekly profit by subtracting C from R, which = P.

I need to show that, in terms of the number of units produced (X), the weekly profit is given by, P =140X - 500 - 3X^2.

So far I havedone P = R - C

p = (150X - 3x^2) -(500 + 10X)
= 140X - 3X^2 - 500

Have i shoen that in terms of X as I needed to, as my 500 and 3X^2 are the other way round....This is what confuses me.

Any light you could shed on this would be great,
Thank you.

2. Its the same. Rewrite if you like. Changing the order of the terms does not change the answer.

140X-3X^2-500=140X-500-3X^2

3. Ok, thanks i get that now, but Im still wondering if, as im supposed to show in terms of x, that I need to start the answere of with X=...... instead of P=

4. I would say that, the weekly profit in terms of the number of units produced (X) is P as a function of X. Or short P(X)=.... just like you have done.

It seems to me that this is a fairly simple question with nothing more to it.

Or are there more questions? Like how many units do you have to sell to reach the brake-even-point?

Bye

5. ok thanks alot, I think i was looking a bit too far into it. It also says in the exercises that i have that you can plot a graph of P against X. Would this involve me soving the equation for X=0 ?

6. Hi,

Solving for x=0, means that you have sold zero units. And you would calculate the 'profit' for that amount of units.

Plotting the graph would mean that you calculate the profit for several values of X, units. For instance X=0, X=10, X=20,..., X=100. The calculated values can be put on the y-axis and the X-values (units) on the x-axis.

Have you worked with coordinates? And xy-graphs?

Bye

7. Yeah, i remember how now, I've found some examples to work off. Thanks again.