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25 October, 18:39

Maximize Q=3xy^2, where x and y are positive numbers such that x+y^2=2

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  1. 25 October, 18:47
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    Sub x = 2-y^2 to Q, we get:

    Q = 3 (2-y^2) * y^2

    let y^2 = k

    Q = 3 (2-k) k = 3 (2k-k^2)

    2k-k^2 has a max when k = 1

    Then y^2 = 1 - > y = 1 or - 1
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