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Math Help - Finding Value of x at which y is a maximum

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    Finding Value of x at which y is a maximum

    y = -x^2 + 20x

    Finding Value of x at which y is a maximum. What is the maximum value of Y.

    This is obviously so easy but I need help because Im stupid and suffering from a severe mind blank.....In this circumstance will someone just point me in the right direction telling me what i need to do without doing it...obviously involves differentiation and possibly the second derivative
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  2. #2
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    Quote Originally Posted by rooney View Post
    y = -x^2 + 20x

    Finding Value of x at which y is a maximum. What is the maximum value of Y.

    This is obviously so easy but I need help because Im stupid and suffering from a severe mind blank.....In this circumstance will someone just point me in the right direction telling me what i need to do without doing it...obviously involves differentiation and possibly the second derivative
    first of all, you should have a mental picture of the graph of y (a parabola with vertex up)

    using calculus ...

    find y' , set y' = 0 to find the x-value of the vertex (the maximum).

    the maximum value is the value of y at that value of x.


    using algebra ...

    x-value of the vertex = -b/(2a)

    using calculus
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    Take the derivative and solve for x. Which would be your relative max or relative min. Then taking the 2nd derivative and if it is f''>0 you have a local minimum. if f''<0 you have a local maximum and if f''=0 then you have a possible inflection point.
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  4. #4
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    ok

    so

    -x^2+20x = 0
    becomes -2x + 20 = 0
    therefore the value of x = 10...correct?
    then are you saying i need to sub into orig function which leaves y=300?
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  5. #5
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    Quote Originally Posted by rooney View Post
    ok

    so

    -x^2+20x = 0
    becomes -2x + 20 = 0
    therefore the value of x = 10...correct?
    then are you saying i need to sub into orig function which leaves y=300?
    its 100. -(10)^2 = -100 :P ur mind seems fried, take a break homie.
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  6. #6
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    Quote Originally Posted by rooney View Post
    ok

    so

    -x^2+20x = 0 you do not set the original function equal to 0, just the derivative
    becomes -2x + 20 = 0
    therefore the value of x = 10...correct?
    then are you saying i need to sub into orig function which leaves y=300? that's right (the subbing in part), not the y=300
    now, answer this ... why do you have to set the derivative equal to 0?
    Last edited by skeeter; July 2nd 2010 at 08:23 AM. Reason: did not notice the evaluated y max
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    Quote Originally Posted by jakncoke View Post
    its 100. -(10)^2 = -100 :P ur mind seems fried, take a break homie.
    -2(10)+20 = 0

    therefore = 10.

    Sub 10 into original equation 10^2 + 20(10)= 300

    hmmm.....i dont see how thats = 100?
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  8. #8
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    y = -x^2 + 20x

    y = -(10^2) + 20(10)

    y = -100 + 200 = 100
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    -10^2 = 100!
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  10. #10
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    no ... -x^2 = -1 \cdot x^2
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    -10^2 = -10 * -10 = 100
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  12. #12
    Senior Member jakncoke's Avatar
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    Quote Originally Posted by rooney View Post
    -10^2 = -10 * -10 = 100
    then what sense would it make to have a -x^2? instead of x^2 ?  -x^2 = -1 * x * x ....  -1 * 10 * 10 = -100 ....
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  13. #13
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    Quote Originally Posted by rooney View Post
    -10^2 = -10 * -10 = 100
    you are confusing -x^2 with (-x)^2 ... the two expressions are different. remember the order of operations?
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  14. #14
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    You don't need calculus to find the maximum value of a quadratic function: complete the square.

    y= -x^2+ 20x= -(x^2- 20x)= -(x^2- 20x+ 100- 100)= -(x- 10)^2+ 100.

    If x= 10, x- 10= 0 so y= 100. If x is any other value, (x- 10)^2 is positive so -(x- 10)^2 is negative and -(x-10)^2+ 100 is less than 100.
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