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calculus

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given y = 4x-3

find the minimum value of xy.
find the rate of change of xy with respect to x.
find the rate of change of xy with respect to y.

  • calculus -

    f(x,y) = xy = x (4x-3)
    f = 4 x^2 - 3x
    df/dx = 8 x - 3
    that is 0 when x = 3/8 and the second derivative is positive so it is a minumum
    then x y = (3/8)(3/2 -3)
    = (3/8) (-3/2) = -9/16

    we found df/dx

    now df/dy
    4x = y+3
    x = (y+3)/4
    g(y) = xy = y(y+3)/4
    = (1/4)(y^2+3y)
    dg/dy = (1/4)(2 y+3)
    = y/2 + 3/4

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