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Q=3x^2+4y^2 if x+y=7 what is minimum value for Q.
I have absolutely no idea how to solve this problem.

  • math -

    from the 2nd equation: y = 7-x
    sub into the other

    Q = 3x^2 + 4(7-x)^2)
    = 3x^2 + 4(49 - 14x + x^2
    = -x^2 + 196 - 56x + 4x^2
    = 3x^2 - 56x + 196

    The minimum of Q occurs at the vertex of the corresponding parabola

    If you know Calculus ...
    dQ/dx = 6x - 56 = 0 for a min of Q
    x = 56/6 = 28/3

    Q = 3(28/3)^2 - 56(28/3) + 196 = -196/3

    If no Calculus, then complete the square ...
    Q = 3(x^2 - (56/3)x + 784/9 - 784/9) + 196
    = 3( (x - 28/3)^2 - 784/9) + 196
    = 3(x-28/3)^2 - 784/3 + 196
    = 3(x-28/3)^2 - 196/3

    so the min of -196/3 happens when x = 28/3

  • math -

    What is the distance between the parallel lines, whose equation are 3x + 4y = 2 and 3x + 4y = -5.

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