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Homework Help: Maximum and minimum values

  1. Jun 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Show that the maximum and minimum values of y occurs when x^3=8+- 2(14)^1/2

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 21, 2014 #2

    PhysicoRaj

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    What have you attempted at? Do you know the method to solve this?

    Plus check this: x^3=8+- 2(14)^1/2

    is it : ##x^3 = 8+(-2)(14)^{1/2}## ??
     
  4. Jun 21, 2014 #3

    PhysicoRaj

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    What is y defined to be?
     
  5. Jun 21, 2014 #4
    It should be +on the top and minus on the bottom,not sure how to write it and I have been using the sign test to test it,however the sign test should be for y values,but only x values are given by the question,plus do I sub the x values into the y^3 equation to find y or should I use dy/dx =(2y-x^2)/(y^2-2x)=0 then sub x values to find y?
    -x^2)
     
  6. Jun 21, 2014 #5

    PhysicoRaj

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    You should have definition for y, right? Else what would you differentiate?
     
  7. Jun 21, 2014 #6
    Please post the full question.
     
  8. Jun 21, 2014 #7
    Y^3=6xy-x^3-1
     
  9. Jun 21, 2014 #8
    I think I have posted the attachment now,sorry I am a first timer here
     

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  10. Jun 21, 2014 #9

    PhysicoRaj

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    Okay,

    ##y^3=6xy-x^3-1##

    and

    ##x^3=8\pm2\sqrt14##

    right?

    Now how have you decided to proceed?
     
  11. Jun 21, 2014 #10
    Yup correct
     
  12. Jun 21, 2014 #11

    PhysicoRaj

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    Well, that completely makes sense now. What have you attempted?
     
  13. Jun 21, 2014 #12
    As I know at stationary point,dy/dx=0 so from dy/dx=(2y-x^2)/(y^2-2x),from there I can find y=(x^2)/2,then I substitute x=2+2(14)^(1/6) and x=2-2(14)^(1/6) to get y values right? Then I reciprocate the dy/dx =dx/dy and then I do the sign test? But I got all + signs which means I fail to prove y has maximum and minimum values.
     
  14. Jun 21, 2014 #13

    PhysicoRaj

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    Right.

    Ok that will fetch you the maximum and minimum of y.

    I'm not sure if I understood this one.

    But as soon as you get the ##y=x^2/2##, can you use it and the 'given' to solve for values of x?
     
  15. Jun 21, 2014 #14

    PhysicoRaj

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    Note: the question is to show that the maximum and minimum of y occurs at x=so and so. Not to find the maximum and minimum values of y itself. So, you have to arrive at the x values, rather than substituting them.
     
  16. Jun 21, 2014 #15
    I haven't attempt which I didn't thought of it that but as I sub it back into y^3, I get x=2 and y=2,so now I have 3 values of x and I can find 3 values of y,and then I need to sub into the second derivative test or using a sign test but If I do a sign test there,there are two variables,how to do it ? @@
     
  17. Jun 21, 2014 #16
    Oo then by using the 3 values of x,I do a second derivative test and sub the values in order to prove that the x values are maximum and minimum values and we can show that maximum and minimum value of y occurs at those x right? And can I do a sign test to test whether y is decreasing or increasing?
     
  18. Jun 21, 2014 #17

    PhysicoRaj

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    You mean you substituted y=x^2/2 in the original y equation. What did it yield? Which are the three x's and y's? :confused:
     
  19. Jun 21, 2014 #18
    Yup I sub back into the original y equation and get another value of x and y ,however I just noticed that I don't actually need it right? Because I can do a 2nd derivative and put the previous x and y values into it and find whether it is maximum or minimum values,but again may I use a sign test for an implicit differentiation equation to find maximum and values oo?
     
  20. Jun 21, 2014 #19

    PhysicoRaj

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    No, if you can find the points of maxima and minima without using the maximum and minimum values, it's alright then. But if you require them to solve for the points of inflection, you will have to find the max. and min. values.
     
  21. Jun 21, 2014 #20

    PhysicoRaj

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    Can you post what you actually worked out? Because it's difficult to imagine what you have done or find the point where you are lacking.
    If typing in was difficult, a snapshot of your work will do.
     
  22. Jun 21, 2014 #21
    I just tried the 2nd derivative

    I have used the calculator to change the x^3 values into 15.84 and -5.84,but when I substitute the x values and y values (how I find y values,I get it from y=(x^2)/2,substituting the x values as stated above) into my second derivative but in the end I got all positive value which means only local minimum values only. So I did it wrongly?@@
     

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  23. Jun 21, 2014 #22
    This is another method that I attempted to do but it is wrong
     

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  24. Jun 21, 2014 #23

    PhysicoRaj

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    How did you get this? Probably by the previous x^3 equation which they have given?

    The implication is right, but I don't know if all are minima or maxima or both, because I haven't worked it out. [STRIKE]If all are minima, that doesn't have anything to do with the answer they expect.[/STRIKE] Okay I saw what the question asks for.
     
  25. Jun 21, 2014 #24

    PhysicoRaj

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    I think they expect you to arrive at ##x^3=8+-2(14)^{1/2}## rather than assume it at the first. This is what is going wrong.
     
  26. Jun 21, 2014 #25
    Yeah,the x values are from the x^3 that they provide,when you substitute those values,you will find that there are many zero values so it is able to get the d2y/dx2= +value which is greater than zero so both values of x are local minimums
     
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