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Math Help - Maximize

  1. #1
    Junior Member
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    Post Maximize

    Hi, I need some help with this problem:

    Maximize Revenue given the following: R(x,y)=x(1000-0.1x2)+y(200-0.02y) subject to x>0, y>0, and 8x+3y<2400.

    This is what I have so far but I don’t know what to do next.
    Can some one help me please

    ▼R(x,y)=(1000-0.3x2)i+(200-0.04y)j
    1000-0.3x2=0 --> x = ±√(1000/0.3)
    200-0.04y=0 --> y = ±√(200/0.04)
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by MarianaA View Post
    Hi, I need some help with this problem:

    Maximize Revenue given the following: R(x,y)=x(1000-0.1x2)+y(200-0.02y) subject to x>0, y>0, and 8x+3y<2400.

    This is what I have so far but I don’t know what to do next.
    Can some one help me please

    ▼R(x,y)=(1000-0.3x2)i+(200-0.04y)j
    1000-0.3x2=0 --> x = ±√(1000/0.3)
    200-0.04y=0 --> y = ±√(200/0.04)
    The second of these should be y=200/0.04=5000. Which gives the
    critical points at (57.735,5000) and (-57.735,5000). Neither of these
    are in the feasible region.

    As the constraints are of strict inequality type there is no point in
    the faesible region at which R(x,y) takes a maximal value.

    If the constraints were:

    x>=0, y>=0, and 8x+3y<=2400,

    then the maximum would occur on the boundary of the feasible region, and
    one could find it by examining R(x,y) on the boundary.

    RonL
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  3. #3
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    Quote Originally Posted by CaptainBlack View Post
    The second of these should be y=200/0.04=5000. Which gives the
    critical points at (57.735,5000) and (-57.735,5000). Neither of these
    are in the feasible region.

    As the constraints are of strict inequality type there is no point in
    the faesible region at which R(x,y) takes a maximal value.

    If the constraints were:

    x>=0, y>=0, and 8x+3y<=2400,

    then the maximum would occur on the boundary of the feasible region, and
    one could find it by examining R(x,y) on the boundary.

    RonL
    but it is y = ± \sqrt(200/0.04) not y=200/0.04
    and how you get 57.735
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  4. #4
    Junior Member
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    Quote Originally Posted by CaptainBlack View Post
    The second of these should be y=200/0.04=5000. Which gives the
    critical points at (57.735,5000) and (-57.735,5000). Neither of these
    are in the feasible region.

    As the constraints are of strict inequality type there is no point in
    the faesible region at which R(x,y) takes a maximal value.

    If the constraints were:

    x>=0, y>=0, and 8x+3y<=2400,

    then the maximum would occur on the boundary of the feasible region, and
    one could find it by examining R(x,y) on the boundary.

    RonL
    ok I undestand how you get 57.735, you get it by dividing \sqrt(200/0.04) and 5000 by dividing 200/.04
    But why?
    Dont I need to substitute the values x = ± \sqrt(1000/0.3) , y = ± \sqrt(200/0.04) the equation 8x+3y<2400
    if it is this is what i get
    8( \sqrt(1000/0.3) )+3( \sqrt(200/0.04) )<2400 = 674.01

    8(- \sqrt(1000/0.3) )+3( \sqrt(200/0.04) )<2400 = -249.74

    8( \sqrt(1000/0.3) )+3(- \sqrt(200/0.04) )<2400 = 249.74

    8(- \sqrt(1000/0.3) )+3(- \sqrt(200/0.04) )<2400 = -674.01
    Last edited by MarianaA; October 25th 2007 at 06:28 AM.
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