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Math Help - Max / Min

  1. #1
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    Smile Max / Min

    Hello

    I had following exercise at my exam last wednesday:

    Find the nominal value and the place of the maxmum and the minimum of the function f(x,y,z) = x + y + z^2
    on the area of the ball with (x^2 + y^2 + z^2 <= 1) ....
    can please someone help me. just wanna check if i did it right...
    thank you guys!!!

    ps: sorry if my expressions arn't correct... hope you understand what i mean tho.... im swiss
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  2. #2
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    Quote Originally Posted by mrintosh View Post
    Hello

    I had following exercise at my exam last wednesday:

    Find the nominal value and the place of the maxmum and the minimum of the function f(x,y,z) = x + y + z^2
    on the area of the ball with (x^2 + y^2 + z^2 <= 1) ....
    can please someone help me. just wanna check if i did it right...
    You need to optimize f(x,y,z) = x+y+z^2 subject to g(x,y,z)=1 with g(x,y,z)=x^2+y^2+z^2.

    That means,
    \left\{ \begin{array}{c}\nabla f = k \nabla g \\ x^2+y^2+z^2 = 1 \end{array} \right.

    Now you need to solve for (x,y,z).
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  3. #3
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    well, i looked first if it there is a max/min in the inside of the ball:

    fx = 1
    fy = 1
    fz = 2z

    it has to be: fx=fy=fz=0 but thats notpossible, so there is no max/min in the inside.

    outside (surface):

    transform x, y, z into spherical-coordinats:

    x = sin(p)cos(q)
    y = sin(p)sin(q)
    z = cos(p)

    than i wrote the function (f=x+y+z^2) new:
    f(x(p,q), y(p,q), z(p,q) = .....

    and than i did: df/dp and df/dq

    df/dp=df/dq=0 for max/min, so i got for the value +-√2

    and for the place maxx,y,z)=( (√2)/2 , (√2)/2 , 0)
    min: ( -(√2)/2 , -(√2)/2 , 0)

    is that right???? please say yes buddy
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  4. #4
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    Quote Originally Posted by mrintosh View Post
    Hello

    I had following exercise at my exam last wednesday:

    Find the nominal value and the place of the maxmum and the minimum of the function f(x,y,z) = x + y + z^2
    on the area of the ball with (x^2 + y^2 + z^2 <= 1) ....
    can please someone help me. just wanna check if i did it right...
    thank you guys!!!
    I did a different problem, I maximized f(x,y,z) on the boundary.

    But you wanted to maximize the boundary including the interior. The boundary can be done with Lagrange Multipliers like in my first post. The interior is found by doing f_x=f_y=f_z=0. Which as you said is impossible. Hence look at the boundary.
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  5. #5
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    and what do you think, is my solution correct (the value and place)???
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  6. #6
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    "But you wanted to maximize the boundary including the interior"

    well no, i only looked at the surface:

    usually: x= Rsin(p)cos(p) -> i made R=1 -> its only the suface, isnt it???
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  7. #7
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    Quote Originally Posted by mrintosh View Post
    and what do you think, is my solution correct (the value and place)???
    No. Here is a non-Calculus approach.

    \mbox{max}_{x^2+y^2+z^2=1}\left\{ x+y+z^2 \right\} = \frac{3}{2}.

    Because,
    x+y+z^2 = x+y+(1-x^2-y^2) = \frac{3}{2} - \left( x + \frac{1}{2} \right)^2  - \left( y+\frac{1}{2} \right)^2. So the maximum is \frac{3}{2} \mbox{ when }x=-\frac{1}{2},y=-\frac{1}{2},z=\pm \frac{\sqrt{2}}{2}.
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  8. #8
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    ****!!!!! you're right!
    well thank you so much for your help!!!
    next time i will do it right...
    cheers
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  9. #9
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    Quote Originally Posted by mrintosh View Post
    Hello

    I had following exercise at my exam last wednesday:

    Find the nominal value and the place of the maxmum and the minimum of the function f(x,y,z) = x + y + z^2
    on the area of the ball with (x^2 + y^2 + z^2 <= 1) ....
    can please someone help me. just wanna check if i did it right...
    thank you guys!!!

    ps: sorry if my expressions arn't correct... hope you understand what i mean tho.... im swiss
    Lagrange multipliers.

    RonL
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  10. #10
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    Quote Originally Posted by CaptainBlank View Post
    Lagrange multipliers.

    RonL
    Look at my first post.
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