word problem

12 is divided into two parts such that the product of the square of one part and fourth power of the other give a maximum,find the two numba

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  1. x+y=12
    z = x^2 y^(1/4)
    = (12-y)^2 y^1/4
    = 144y^1/4 - 24 y^5/4 + y^9/4

    dz/dy = 36/y^3/4 - 30y^1/4 + 9/4 y^5/4
    = (36 - 30y + 9/4 y^2) / y^3/4
    dz/dy=0 when
    9y^2-120y+144 = 0
    3(3y-4)(y-12) = 0
    clearly, y=12 gives z=0, a minimum
    y=3/4 gives x=45/4 and z=(1024/9) 4throot(4/3)

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