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Math Help - Lagrange Multipliers

  1. #1
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    Lagrange Multipliers

    I'm supposed to find the maximum value of f(x,y)=xy if (x,y) is a point on the ellipse 9x^2+16y^2=144.

    I have that y=lambda18x and x=lambda32y but I'm not sure how to go about solving for x and y.

    Thanks
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by jelloish View Post
    I'm supposed to find the maximum value of f(x,y)=xy if (x,y) is a point on the ellipse 9x^2+16y^2=144.

    I have that y=lambda18x and x=lambda32y but I'm not sure how to go about solving for x and y.

    Thanks
    \nabla(xy) = \langle y,x\rangle and \nabla(9x^2+16y^2) = \langle 18x,32y\rangle

    So \langle y,x\rangle = \lambda\langle 18x,32y\rangle

    So you want to solve the system:
    y=18\lambda x
    x=32\lambda y
    9x^2+16y^2=144

    The best thing to do is to get rid of \lambda first. So divide the first two equations to get: \frac{x}{y} = \frac{16y}{9x} \implies 9x^2=16y^2

    Substituting into the third equation gives us 9x^2+9x^2=144 \implies x=2\sqrt{2} \implies y=\sqrt{\frac{9}{2}}=\frac{3\sqrt{2}}{2}

    So the maximum of xy occurs when x=2\sqrt{2} and y=\frac{3\sqrt{2}}{2}, so the maximum value is 2\sqrt{2}\cdot\frac{3\sqrt{2}}{2} = \boxed{6}
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  3. #3
    Senior Member Spec's Avatar
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    Another way to solve it is to switch to polar coordinates: 3x=12cos(t), 4y=12sin(t)

    Then you just need to find the maximum of a function of one variable:
    f(x,y)=xy=f(4cos(t),3sin(t))=12cos(t) sin(t)
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by Spec View Post
    Another way to solve it is to switch to polar coordinates: 3x=12cos(t), 4y=12sin(t)

    Then you just need to find the maximum of a function of one variable:
    f(x,y)=xy=f(4cos(t),3sin(t))=12cos(t) sin(t)
    True. The derivative of 12\cos(t)\sin(t) is 12\cos(2t). Setting 12\cos(2t)=0 gives you t=\frac{\pi}{4}.

    12\cos(\pi/4)\sin(\pi/4) = 12\cdot\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2}}{2} = \boxed{6}

    which is the same answer as above.
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