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mathslover

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

  • one year ago
  • one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    • one year ago
  15. DLS
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    should work

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

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

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

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

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

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

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

    • one year ago
  23. DLS
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    @agent0smith

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

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

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

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

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

    • one year ago
  29. Yahoo!
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    Maximum is infinite

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

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

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

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

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

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

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

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

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

    • one year ago
  39. agent0smith
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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

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

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

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

    • one year ago
  43. Yahoo!
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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

    • one year ago
  44. agent0smith
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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.

    • one year ago
  45. agent0smith
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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

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

    • one year ago
  47. DLS
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    @RnR @shubhamsrg

    • one year ago
  48. AravindG
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    yep @DLS

    • one year ago
  49. waterineyes
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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??

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

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

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

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

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

    • one year ago
  55. AravindG
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    for ?

    • one year ago
  56. DLS
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    nothing XD

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

    • one year ago
  58. DLS
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    :")

    • one year ago
  59. klimenkov
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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.

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

    • one year ago
  61. mathslover
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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.

    • one year ago
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