Lagrange Multipliers-Calculus 3?

Find the shortest distance between the point (0,0) and the line y=x+1 using Lagrange Multipliers.

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  • ted s
    Lv 7
    7 years ago
    Favorite Answer

    So you desire to minimize D = √ ( x² + y² ) subject to 0 = x - y + 1

    let F = √(x² + y² ) + µ ( x - y + 1)...compute δF/δx and δF / δy , set both to 0 and solve..

    then find D................................. { 1 / √2 }

  • hatti
    Lv 4
    4 years ago

    i visit offer it a attempt, yet i do now not comprehend if my answer may be appropriate. 2xy^2z^2 = 2x? 2yx^2z^2 = 2y? 2zx^2y^2 = 2z? x^2+y^2+z^2 = a million xy^2z^2 = x? yx^2z^2 = y? zx^2y^2 = z? x^2+y^2+z^2 = a million Multiply equation a million by potential of x, and equation 2 by potential of y xy^2z^2 = x? x^2y^2z^2 = x^2? yx^2z^2 = y? x^2y^2z^2 = y^2? Subtract the recent equations 0 = x^2? - y^2? component out ? 0 = ?(x^2 - y^2) this is not acceptable what ? is, through fact we are in a position to divide that out. If ? = 0, then x,y,z = 0, which does now not fulfill the regulations. So 0 = ?(x^2 - y^2) = (x+y)(x-y) Giving x = -y, and x = y Multiply equation 2 by potential of y, and equation 3 by potential of z and subtract, giving y^2? - z^2? = 0 ?(y^2 - z^2) = 0 ?(y-z)(y+z) = 0 So y = z, and y = -z ---> y = z, -y = z And x = -y = z, and x = -y = z. Plugging in... x^2 + y^2 + z^2 = a million x^2 + (-x)^2 + x^2 = a million 3x^2 = a million x^2 = a million/3 x = (a million/3)^(a million/2), and -(a million/3)^(a million/2) and so on. Edit: pass with kb's answer, he's helped me a lot with my own calc questions!

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