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lamthornhill

  • 2 years ago

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

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  1. richyw
    • 2 years ago
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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?

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

  3. Sheng
    • 2 years ago
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    |dw:1352089616943:dw|

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

  5. Sheng
    • 2 years ago
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    b = -5 sorry

  6. richyw
    • 2 years ago
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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

  7. lamthornhill
    • 2 years ago
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    I have no idea what you are talking about

  8. richyw
    • 2 years ago
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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.

  9. richyw
    • 2 years ago
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    i'm getting there!

  10. lamthornhill
    • 2 years ago
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    ok

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

  12. richyw
    • 2 years ago
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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.

  13. richyw
    • 2 years ago
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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?

  14. richyw
    • 2 years ago
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    sorry \(c=-8850\)

  15. richyw
    • 2 years ago
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    now plug those numbers into the the quadratic equation.

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

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

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

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

  20. richyw
    • 2 years ago
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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.

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

  22. lamthornhill
    • 2 years ago
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    ok

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

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

  25. richyw
    • 2 years ago
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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)

  26. lamthornhill
    • 2 years ago
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    ok

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

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

  29. richyw
    • 2 years ago
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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...

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

  31. lamthornhill
    • 2 years ago
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    for the top I came up 34225

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

  33. lamthornhill
    • 2 years ago
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    ok... I'm not good with math.....

  34. richyw
    • 2 years ago
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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

  35. richyw
    • 2 years ago
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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).

  36. lamthornhill
    • 2 years ago
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    ty

  37. richyw
    • 2 years ago
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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.

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