Price vs Revenue equation

I think I have found the right answer but if someone could have a look at it for me, that would be awesome. Thanks.

The John Deere company has found that the revenue from sales of heavy-duty tractors is a function of the unit price *p,* in dollars, that it charges. The revenue *R, *in dollars is -.05p^2 + 1900p

At what prices p is revenue zero? Round to four decimal places. ** I have when revenue is zero, then p=0 or 3800. (0,0) and (3800,0)**

For what range of prices will revenue exceed $1,300,000? Round to four decimal places.

**886**__<__p__<__2904 The price must be between or equal to the values of $886 and $2900 for revenue to exceed $1,300,000.

For this part, I just looked on the table of my calculator. Should I be more precise and use a formula?

Re: Price vs Revenue equation

I'm getting something a little different. Is this the correct equation?

$\displaystyle R = -0.05p^2 + 1900p.$

The second root should be $\displaystyle p=38\,000$ rather than 3800. For the second part, you can get an exact solution by solving the quadratic

$\displaystyle -0.05p^2 + 1900p = 1\,300\,000.$

My bounds come out differently than yours.

Re: Price vs Revenue equation

My mistake. it is .5 instead of .05

With that being the case, it looks like I am right then. 1900/.5 is 3800 and 1900/.05 is 38,000

I added the extra zero so that gave you an extra zero.

The second part is what is messing with me. The way it states, I need to find what p values (or x) will make the R values (or y) exceed $1,300,000. Looking at the table, that that is 886 to 2900. I am unsure how to get those values via the equation.

I know I need to do -.5p^2+1900p > 1,300,000 but I am kinda lost from there. Should I just move the 13m over with everything else and then factor it?

Re: Price vs Revenue equation

Quote:

Originally Posted by

**wevie** I know I need to do -.5p^2+1900p > 1,300,000 but I am kinda lost from there. Should I just move the 13m over with everything else and then factor it?

Yes. Do you know how to solve a quadratic equation? I don't think this will factor nicely, so you will probably have to use the quadratic formula.

$\displaystyle -0.5p^2+1900p-1\,300\,000 = 0$

The two roots of this equation will be the endpoints of your interval.

Re: Price vs Revenue equation

Yes. I know how to factor. I came up with 895.0124 and 2904.9876

So am I correct in that I should right it 895.0124__<__p__<__2904.9876 ?