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Thread: max and min pre cal

  1. #1
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    Question max and min pre cal

    I can't seem to calculate the min of this problem.
    c=350,000-140x+0.030xsq
    Could you please walk me through the problem?
    thanks
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Donnastar View Post
    I can't seem to calculate the min of this problem.
    c=350,000-140x+0.030xsq
    Could you please walk me through the problem?
    thanks
    $\displaystyle c=.030x^2-140x+350000$

    Find the x-coordinate of the vertex using $\displaystyle x=\frac{-b}{2a}$

    Now, find $\displaystyle c\left(\frac{-b}{2a}\right)$
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  3. #3
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    Question max and min

    I can't come up with the correct answer when I pulg the numbers into the formula. Could you make it real simple for me? thanks
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  4. #4
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    Quote Originally Posted by Donnastar View Post
    I can't come up with the correct answer when I pulg the numbers into the formula. Could you make it real simple for me? thanks
    $\displaystyle c=ax^2+bx+c$

    $\displaystyle c=.030x^2-140x+350000$

    $\displaystyle a=.030$

    $\displaystyle b=-140$

    $\displaystyle c=350000$

    Substituting,

    $\displaystyle x=\frac{-b}{2a}=\frac{140}{.06}=\frac{7000}{3}$

    $\displaystyle c\left(\frac{7000}{3}\right)=.030\left(\frac{7000} {3}\right)^2-140\left(\frac{7000}{3}\right)+350000$

    $\displaystyle c\left(\frac{7000}{3}\right)=\boxed{\frac{560000}{ 3}}$

    That should be your minimum.
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