Maximums and minimums

  1. 1. The problem statement, all variables and given/known data

    y3=6xy-x3-1 , find dy/dx . Prove that the maximm value of y occurs when
    x3=8+2sqrt(114) and the minimum value when x3=8-2sqrt(114)

    2. Relevant equations



    3. The attempt at a solution

    I found dy/dx to be (2y-x2)/(y2-2x)

    Then from here , dy/dx=0 for turning points and it happens when y=1/2 x2

    Substitute this back to the original equation and find x from there using the quadratic formula

    [tex]x^3=\frac{16\pm\sqrt{16^2-4(1)(8)}}{2}[/tex] which simplifies to

    [tex]x^3=8\pm \sqrt{56}[/tex]

    First off , the x coordinate of my turning points is wrong but where is that mistake ? Also , How do i test whether this is a max or min ?

    Second order differentiation doesn't help ..
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,043
    Science Advisor
    Homework Helper

    Hi thereddevils! :smile:

    (have a square-root: √ :wink:)
    Your solution looks correct to me.

    The one in the book has √456 instead of √56 … perhaps someone pressed the wrong button?
    It should do … just differentiate your equation for dy/dx wrt x, and remember you can put dy/dx = 0. :wink:
     
  4. thanks Tiny ! I am still a little confused with the max and mins .

    I am a bit reluctant to do the second order differentiation as there are 'y's in the dy/dx equation .

    When i put dy/dx=0 , y= 1/2 x^2 can i substitute this into the dy/dx equation to get rid of the y ?
     
  5. tiny-tim

    tiny-tim 26,043
    Science Advisor
    Homework Helper

    Yes, if you already have dy/dx = f(x,y), then d/dx it to get

    d2y/dx2 = ∂/∂x(f(x,y)) + ∂/∂y(f(x,y))dy/dx

    the second term will be zero, and you put the x = 3√(8 ± etc) and y = 1/2 x2 into the first term. :smile:
     
  6. erm , that looks to be partial differentiation and i am not so familiar with that but i would like to learn that :biggrin:

    Can i differentiate implicitly instead of partial diff ?

    I tried ,

    [tex]\frac{d^2y}{dx^2}=\frac{(y^2-2x)(2dy/dx-2x)-(2y-x^2)(2y dy/dx-2)}{(y^2-2x)^2}[/tex]

    when dy/dx=0 , y= 1/2 x^2

    d2y/dx2=-8/(x3-8)

    x= 2.49(max) or 0.80(min)
     
    Last edited: May 2, 2010
  7. tiny-tim

    tiny-tim 26,043
    Science Advisor
    Homework Helper

    Yes, that's fine (though I haven't checked the last two lines). :smile:
     
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