Asked by Ann
A piece of equipment has cost function C(x)=50x^2 + 1000 and its revenue
function is R(x)= 500x - x^2, where x is in thousands of items. What is the
least number of items that must be sold in order to break even?
50x^2 + 1000 = 500x - x^2
50x^2 + x^2 + 500= 0
51x^2 + 500 = 0
Do I just do -b/2a = -500/102= -250/51? What do I do next?
function is R(x)= 500x - x^2, where x is in thousands of items. What is the
least number of items that must be sold in order to break even?
50x^2 + 1000 = 500x - x^2
50x^2 + x^2 + 500= 0
51x^2 + 500 = 0
Do I just do -b/2a = -500/102= -250/51? What do I do next?
Answers
Answered by
Steve
well, you should see right off that since 51x^2 is always positive, you can never have 51x^2 = -500.
So, let's see what happened.
50x^2 + 1000 = 500x - x^2
51x^2 - 500x + 1000 = 0
Now just use the quadratic formula to find the roots.
So, let's see what happened.
50x^2 + 1000 = 500x - x^2
51x^2 - 500x + 1000 = 0
Now just use the quadratic formula to find the roots.
Answered by
Ann
wow that was a silly mistake I didn't see the x next to 500 and thought I could just subtract it from 1000
thank you!
thank you!
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