Cost and revenue - break even points

kickingtoad

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Nov 12, 2010
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The total revenue and total cost functions for the production and sale of x TV's are given as

\(\displaystyle {R(x)=190x-0.2x^{2}}\)
\(\displaystyle {C(x)=3550+24x}\)

List the values of x at the break even point(s). It is possible that there are no break even points.

I think to get the break even points you subtract C(x) from R(x) and set it equal to 0.

\(\displaystyle {R(x)-C(x)=0}\)

\(\displaystyle {190x-0.2x^{2}-(3550+24x)=0}\)

\(\displaystyle {-0.2x^{2}+166x-3550=0}\)

Where do I go from here?
 
This is not a calculus problem. It is being moved to Intermediate Algebra.

Use the quadratic formula to solve for x. There will be two values.
 
Ok.

I need to solve this. I got the equation from above from R(x)-C(x)=0 and need to find the break even points.

\(\displaystyle {-0.2x^{2}+166x-3550=0\)

I need to use the quadratic formula. I'm not getting this right though... What am I doing wrong?

Quadratic formula
\(\displaystyle \frac{x=-b\pm\sqrt{b^{2}-4ac}}{2a}\)

1. \(\displaystyle \frac{-166\pm\sqrt{166^{2}-4(-0.2)(-3550)}}{2(-0.2)}\)

2. \(\displaystyle \frac{-166\pm\sqrt{27556-2840}}{-0.4}\)

3. \(\displaystyle \frac{-166\pm\sqrt{24716}}{-0.4}\)

This comes out as incorrect. What is wrong?
 
You have both solutions. One is 21.97

The other is 808.03.

You do know that in the quadratic formula there are two solutions?. Hence, the plus and minus cases.
 
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