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mathslover

  • one year ago

How to find the maximum and minimum value of the function : \(y = 25x^2 + 5 - 10x\)

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  1. mathslover
    • one year ago
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    Well I took : \(\cfrac{dy}{dt} = 0\) And then: \(\cfrac{d(25x^2+5-10x)}{dt}= 0\) \(\implies 50x - 10 = 0\) \(\implies 50x = 10\) \(\implies x = \cfrac{10}{50} = \cfrac{\cancel{10}^1}{\cancel{50}^5} = \cfrac{1}{5} \)

  2. DLS
    • one year ago
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    Find the second derivative now.

  3. mathslover
    • one year ago
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    therefore y has minimum value at x = 1/5 What about maximum.

  4. DLS
    • one year ago
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    You have to find 2nd derivative for maximum value

  5. DLS
    • one year ago
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    \[\LARGE y=50x-10\] diff this.

  6. DLS
    • one year ago
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    You should get y=50.

  7. DLS
    • one year ago
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    That is the maximum value

  8. DLS
    • one year ago
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    As far as I remember,you can always check it though.

  9. mathslover
    • one year ago
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    \(\cfrac{d(50x-10)}{dt} \implies y = 50\)

  10. mathslover
    • one year ago
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    Thanks @DLS got it. Is their any condition for negative or positive?

  11. DLS
    • one year ago
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    I think its not an equal sign right there,its > or < I'll let you know.

  12. mathslover
    • one year ago
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    OH : \(\cfrac{d^2 y}{dt^2} = 50\)

  13. DLS
    • one year ago
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    yes :) I wrote, second derivative should be 0

  14. mathslover
    • one year ago
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    Should I substitute x = 1/5 in the equation? to find \(y_{min.}\)

  15. DLS
    • one year ago
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    should work

  16. mathslover
    • one year ago
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    \(y_{min.} = 25(\cfrac{1}{5})^2 + 5 - 10(\cfrac{1}{5})\) \(y_{min.} =\cancel{25}^1(\cfrac{1}{\cancel{25}^1}) + 5 - \cancel{10}^2 (\cfrac{1}{\cancel{5}^1}\) \(y_{min.} = 6-2 = 4\)

  17. DLS
    • one year ago
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    Seems correct,do you have the answer?

  18. mathslover
    • one year ago
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    Hmn ... let me check the answers.

  19. mathslover
    • one year ago
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    Only minimum value = 4 is given ... what would be maximum value?

  20. mathslover
    • one year ago
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    Max. value is not given , might be author forgot that.

  21. mathslover
    • one year ago
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    Though , can you help me finding max. value ?

  22. DLS
    • one year ago
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    isn't it x=50?

  23. DLS
    • one year ago
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    @agent0smith

  24. Yahoo!
    • one year ago
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    y = 25x^2 +5-10x y' = 50x - 10 y'' = 50 Since y'' > 0 The FUction has minimum value

  25. Yahoo!
    • one year ago
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    Minimum value is obstained bY: y' = 0

  26. Yahoo!
    • one year ago
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    y' = 50x - 10 = 0 x = 1/5 So...put x=1/5 in the equation....

  27. Yahoo!
    • one year ago
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    This Function has Only Minimum Value....

  28. agent0smith
    • one year ago
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    You can do this question without any calculus at all :P

  29. Yahoo!
    • one year ago
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    Maximum is infinite

  30. Yahoo!
    • one year ago
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    Yup..@agent0smith By using Some Rules in Quadratic Equation

  31. DLS
    • one year ago
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    i was thinking the same..maximum vallue can be anything actually..

  32. Yahoo!
    • one year ago
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    Function Attains Maximum or Minimum at x = -b/2a

  33. DLS
    • one year ago
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    @mathslover Maximum value of the function is x>50

  34. mathslover
    • one year ago
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    Oh k. Got it now.

  35. DLS
    • one year ago
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    Now it can be 51..51000..51000000000000000..anything

  36. DLS
    • one year ago
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    I told you i was confused,there isn't an equality sign,but a greater to one,just confirmed.

  37. DLS
    • one year ago
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    and minimum value x<1/5

  38. agent0smith
    • one year ago
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    Minimum value of the function with be when x=1/5, not when x is < 1/5

  39. agent0smith
    • one year ago
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    As @Yahoo! said, Function Attains Maximum or Minimum at x = -b/2a this is x = -(-10)/(2*25) = 1/5, no calculus needed :D

  40. agent0smith
    • one year ago
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    And since a is positive (in ax^2+bx+c) the function is concave up, so it's a minimum.

  41. DLS
    • one year ago
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    \[\LARGE 6x^5+5x^4+4x^3+3x^2+2x+1\] @Yahoo! You got a formula for minimizing this?

  42. Yahoo!
    • one year ago
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    Here u have to use Calculus Since it is not a Quadratic Equation...

  43. Yahoo!
    • one year ago
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    find f'(x) first equate it to 0 and find the values of x then find f''(x) by inserting the values of x u got.. if f''(x) > 0 then function has minimum at that point of x and vice versa

  44. agent0smith
    • one year ago
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    @DLS there aren't general formulas for solving quintics or polynomials with order higher than that, this is the Abel-Ruffini theorem: http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Abel%E2%80%93Ruffini_theorem.html There are formulas for quadratics (obviously), cubics and quartics.

  45. agent0smith
    • one year ago
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    If you'd like to solve a quartic of the form \[\large ax^4+bx^3+cx^2+dx +e = 0 \] then here you go: http://planetmath.org/QuarticFormula Or watch wolfram derive it: http://www.wolframalpha.com/input/?i=a*x%5E4%2Bb*x%5E3%2Bc*x%5E2%2Bd*x+%2Be+%3D+0

  46. DLS
    • one year ago
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    For maxima, y=\[\LARGE \frac{d^2(25x^2+5−10x)}{dx^2}<0\]

  47. DLS
    • one year ago
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    @RnR @shubhamsrg

  48. AravindG
    • one year ago
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    yep @DLS

  49. waterineyes
    • one year ago
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    You are taking derivative with respect to x and I respect that too, but @mathslover I think you are writing dt instead of dx.. Right??

  50. AravindG
    • one year ago
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    just applying second derivative test here ..

  51. AravindG
    • one year ago
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    what c value did you get by the way ?

  52. AravindG
    • one year ago
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    oh 1/5 :)

  53. waterineyes
    • one year ago
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    Hey @yahoo, Congratulations for Green.. @AravindG, Congratulations for Ambassador.. And @mathslover, Congratulations for 98 + Green..

  54. DLS
    • one year ago
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    congratulate me too???

  55. AravindG
    • one year ago
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    for ?

  56. DLS
    • one year ago
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    nothing XD

  57. waterineyes
    • one year ago
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    @DLS, Congratulations for getting 1 medal in this post and now 2... :)

  58. DLS
    • one year ago
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    :")

  59. klimenkov
    • one year ago
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    Write \(y(x)=25x^2-10x+5\) as \((5x-1)^2+4\) and use that this is the sum of two non-negative addends. So it has minimum, when \(5x-1=0\quad\Rightarrow\quad x=\frac15 \). To prove that there is no maximum try to compare \(y(x) \) and \(y(x+1)\) by subtracting: \(y(x+1)-y(x)=(25x^2+50x+25-10x-10+5)-(25x^2-10x+5)=50x+15\). So for large \(x\) this is not bounded. So it has no maximum.

  60. mathslover
    • one year ago
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    Thanks a lot everyone for your help

  61. mathslover
    • one year ago
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    Oh thanks @waterineyes , and sorry for using dt there... I meant dx. Actually I usually come up with problems having dt there na.

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