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lamthornhill

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

  • one year ago
  • one year ago

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  1. richyw
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    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?

    • one year ago
  2. lamthornhill
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    no, that is where I'm having the problem cause I don't understand algebra

    • one year ago
  3. Sheng
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    |dw:1352089616943:dw|

    • one year ago
  4. Sheng
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    quadratic formula, where a = 1, b= 5, c = -8550

    • one year ago
  5. Sheng
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    b = -5 sorry

    • one year ago
  6. richyw
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    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

    • one year ago
  7. lamthornhill
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    I have no idea what you are talking about

    • one year ago
  8. richyw
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    \[x^2-5x+70-8620=0\]\[x^2-5x-8850=0\]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.

    • one year ago
  9. richyw
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    i'm getting there!

    • one year ago
  10. lamthornhill
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    ok

    • one year ago
  11. richyw
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    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 \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • one year ago
  12. richyw
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    so looking at the equation we have \(x^2-5x-8850\), you can pick out the parameters a,b and c.

    • one year ago
  13. richyw
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    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?

    • one year ago
  14. richyw
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    sorry \(c=-8850\)

    • one year ago
  15. richyw
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    now plug those numbers into the the quadratic equation.

    • one year ago
  16. lamthornhill
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    sorry I'm not responding....... i'm writing and studying this so I can understand it

    • one year ago
  17. lamthornhill
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    I don't even know what the quadratic equation is

    • one year ago
  18. richyw
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    yup MEMORIZE the quadratic formula! also do you understand what the \(\pm\) sign means?

    • one year ago
  19. lamthornhill
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    no....... this is all new to me....... I didn't take algebra in high school

    • one year ago
  20. richyw
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    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.

    • one year ago
  21. richyw
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    This is the quadratic FORMULA.\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • one year ago
  22. lamthornhill
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    ok

    • one year ago
  23. richyw
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    the \(\pm\)is just used to save space, it means that this is actually two formulas...haha\[\frac{-b+\sqrt{b^2-4ac}}{2a}\]\[\frac{-b-\sqrt{b^2-4ac}}{2a}\]

    • one year ago
  24. lamthornhill
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    I have been on this type of problem now for 2 hours

    • one year ago
  25. richyw
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    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)

    • one year ago
  26. lamthornhill
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    ok

    • one year ago
  27. richyw
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    right so now just plug the numbers in \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-(-5)\pm\sqrt{(-5)^2-4(1)(-8550)}}{2(1)}\]

    • one year ago
  28. richyw
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    this gives \[x=\frac{5+\sqrt{25+4(8550)}}{2}\]\[x=\frac{5-\sqrt{25+4(8550)}}{2}\]

    • one year ago
  29. richyw
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    For the context of the question though you can't have negative sales, you you only need to plug in the top one...

    • one year ago
  30. richyw
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    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.\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

    • one year ago
  31. lamthornhill
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    for the top I came up 34225

    • one year ago
  32. richyw
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    nope that's not correct, you must have messed up on the calculator. Hand on i'll just check wolfram...

    • one year ago
  33. lamthornhill
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    ok... I'm not good with math.....

    • one year ago
  34. richyw
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    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

    • one year ago
  35. richyw
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    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).

    • one year ago
  36. lamthornhill
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    ty

    • one year ago
  37. richyw
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    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.

    • one year ago
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