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Math Help - maxmium values?

  1. #1
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    maxmium values?

    Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1,

    ive got the x values as



    and



    so and make the maximum values, but how do i show that this is the maximum?
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  2. #2
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    Quote Originally Posted by Student_richard View Post
    make the maximum values, but how do i show that this is the maximum?
    Evaluate the functions at those points and see which one is BIGGER.
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    Hello, Student_richard!

    Find the maximum of f(x,y)\:=\:4xy, subject to the condition x^2+4y^2\:=\:1

    ive got the x values as: . x \,=\,\pm\frac{\sqrt{2}}{2}

    . . and y values: . y \,=\,\pm\frac{\sqrt{2}}{4} . . . . Right . . . good work!

    So \left(\frac{\sqrt{2}}{2},\:\frac{\sqrt{2}}{4}\righ  t) and \left(-\frac{\sqrt{2}}{2},\:-\frac{\sqrt{2}}{4}\right) are the maximums

    but how do i show that they are maximums?

    There must be a test for this (even with Lagrange multipliers),
    . . but I'm not familiar with it.

    But it is clear that x and y must have the same sign
    . . in order for f(x,y) \,=\,4xy to be a maximum.

    That may be sufficient . . . or not.

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  4. #4
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    Quote Originally Posted by Student_richard View Post
    Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1,

    ive got the x values as



    and



    so and make the maximum values, but how do i show that this is the maximum?
    Hello,

    please check your calculations. I've got the same x-values but the y-value of the maxima is +1:

    M_1\left(\frac{\sqrt{2}}{2} , 1 \right)~\text{  and  } ~M_2\left(-\frac{\sqrt{2}}{2} , 1 \right)

    (You only considered the condition without using the equation of the function)

    To show that a certain value is indeed a maximum you have to calculate the 2nd derivative. Plug in the x-value. If f''(x) < 0 then you have a maximum.
    Attached Thumbnails Attached Thumbnails maxmium values?-lemniskate1.gif  
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    cheers for that, but i still dont get the y=1. are you sure about that and can anyone else verify the answer?
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  6. #6
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    Quote Originally Posted by Student_richard View Post
    cheers for that, but i still dont get the y=1. are you sure about that and can anyone else verify the answer?
    Hello,

    your results are: x=\pm\frac{\sqrt{2}}{2}=\pm \sqrt{\frac{1}{2}} and y=\pm \frac{\sqrt{2}}{4}=\pm\sqrt{\frac{1}{8}}

    Now plug in these values into the equation

    f(x,y) = 4xy \Longrightarrow~ f\left(\sqrt{\frac{1}{2}}, \sqrt{\frac{1}{8}}  \right) = 4 \cdot \sqrt{\frac{1}{2}} \cdot \sqrt{\frac{1}{8}} = 4 \cdot \sqrt{\frac{1}{16}} =1<br />
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  7. #7
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    Quote Originally Posted by Student_richard View Post
    Find the maximum of f(x,y)=4xy, subject to the condition x2+4y2=1,
    Not a Lagrange multiplier problem. The maximum of f(x,y) is also a local maximum of (f(x,y))^2, so we want to find the extrema of 4x^2y^2 subject to x^2+4y^2=1. So we want the extrema of g(y)=(1-4y^2)y^2, which we find by setting g'(y)=0 and solving for y.

    Doing this we find that y=\pm \frac{\sqrt{2}}{4} , and so the corresponding x's are x=\pm \frac{\sqrt{2}}{2} and x=\pm \frac{\sqrt{2}}{2} .

    (note in doing this I had to assume that y \ne 0, so if the maximum turns out to be negative we will need to throw in the possibility that (1,0) , (-1,0) give a maximum for f(x,y) of 0)

    This gives four posible points for the maximum  \left(\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{4}\right) , \left(-\frac{\sqrt{2}}{2} ,\frac{\sqrt{2}}{4}\right), \left(\frac{\sqrt{2}}{2} ,-\frac{\sqrt{2}}{4}\right) ,  \left(-\frac{\sqrt{2}}{2} ,-\frac{\sqrt{2}}{4}\right).

    Now we could do some clever things to determine which of these corresponds to the maximum, but the easiest way is just to evaluate f(x,y) at each point and the largest will be the maximum.

    RonL
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  8. #8
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    so and make the maximum values, but to test whether they are the maximum, i would find the second dereative of f(x,y) = 4xy?

    is this the right idea, and how would i find the second derivative?
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