Calculus

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Q. Find the minimum value of Q=x^2y subject to the constraint 2x^2+4xy=294.
Its the derivative method.

  • Calculus -

    from your constraint equation
    4xy = 294-2x^2
    y = (294-2x^2)/(4x)

    so Q = x^2 (294-2x^2)/(4x)
    = 147x/2 - x^3 /2

    dQ/dx = 147/2 - (3/2)x^2 = 0 for a min of Q
    3x^2 /2 = 147/2
    3x^2 = 147
    x^2 = 49
    x= ± 7
    then y = ±7

    if x=7, then y=7
    if x = -7, then y = -7

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