anonymous
  • anonymous
if the cost, C, for manufacturing x units of a certain product is given by C=x^2-5x+70, find the number of units manufactured at a cost of $8620
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
richyw
  • richyw
well \(C(x)=x^2-5x+70\) and you know the cost is $8620 so \[x^2-5x+70=8620\]now solve for x. Do you know how to do that?
anonymous
  • anonymous
no, that is where I'm having the problem cause I don't understand algebra
anonymous
  • anonymous
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anonymous
  • anonymous
quadratic formula, where a = 1, b= 5, c = -8550
anonymous
  • anonymous
b = -5 sorry
richyw
  • richyw
alright well the first thing you usually want to do is write this in general form. To do this you subtract 8620 from both sides
anonymous
  • anonymous
I have no idea what you are talking about
richyw
  • richyw
\[x^2-5x+70-8620=0\]\[x^2-5x-8850=0\]Now you want to find the roots to that. First you want to see if it can be factored, if it can then that's great and if not you need to use the quadratic formuala.
richyw
  • richyw
i'm getting there!
anonymous
  • anonymous
ok
richyw
  • richyw
the quadratic formula finds the roots of quadratic polynomials. For a quadratic polynomial in general form, that is \(f(x)=ax^2+bx+c=0\), the quadratic formulas is \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
richyw
  • richyw
so looking at the equation we have \(x^2-5x-8850\), you can pick out the parameters a,b and c.
richyw
  • richyw
here the \(x^2\) term has nothing in front of it, so \(a=1\), the \(x\) term has a -5 in front of it so \(b=-5\), and \(c=8850\). Does that make sense?
richyw
  • richyw
sorry \(c=-8850\)
richyw
  • richyw
now plug those numbers into the the quadratic equation.
anonymous
  • anonymous
sorry I'm not responding....... i'm writing and studying this so I can understand it
anonymous
  • anonymous
I don't even know what the quadratic equation is
richyw
  • richyw
yup MEMORIZE the quadratic formula! also do you understand what the \(\pm\) sign means?
anonymous
  • anonymous
no....... this is all new to me....... I didn't take algebra in high school
richyw
  • richyw
alright well a quadratic EQUATION is a polynomial equation of the second degree. The initial problem you have is a quadratic equation. The quadratic FORMULA is the formula I gave you. You use the quadratic formula to find the roots of quadratic equations.
richyw
  • richyw
This is the quadratic FORMULA.\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
ok
richyw
  • richyw
the \(\pm\)is just used to save space, it means that this is actually two formulas...haha\[\frac{-b+\sqrt{b^2-4ac}}{2a}\]\[\frac{-b-\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
I have been on this type of problem now for 2 hours
richyw
  • richyw
but this equation is only useful if your equation is in general form. That was the first step that we did (when we subtracted 8620)
anonymous
  • anonymous
ok
richyw
  • richyw
right so now just plug the numbers in \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-8550)}}{2(1)}\]
richyw
  • richyw
this gives \[x=\frac{5+\sqrt{25+4(8550)}}{2}\]\[x=\frac{5-\sqrt{25+4(8550)}}{2}\]
richyw
  • richyw
For the context of the question though you can't have negative sales, you you only need to plug in the top one...
richyw
  • richyw
sorry! how could I have forgotten this, if you were confused before, the quadratic equation gives you x... I'll put in an x= there for you :D. Anyways I don't have a calculator, so plug those two above in.\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
anonymous
  • anonymous
for the top I came up 34225
richyw
  • richyw
nope that's not correct, you must have messed up on the calculator. Hand on i'll just check wolfram...
anonymous
  • anonymous
ok... I'm not good with math.....
richyw
  • richyw
well we already finished the math, now it's just using the calculator :). But the two roots of this polynomial are x=-90 and x=95. Obviously the answer to the question is x=95, because you can't manufacture a negative number of products. Just you know you can always verify this on wolfram alpha http://www.wolframalpha.com/input/?i=x%5E2-5x%2B70%3D8620
richyw
  • richyw
It might seem compplicated but really all you did was two steps 1)Put equation in general form 2)Find the roots (in this case we found the roots using the quadratic equation).
anonymous
  • anonymous
ty
richyw
  • richyw
No worries, and good luck with your math studies! I only learned algebra after high school as well. It just takes a LOT of work! It is sure worth it though.

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